4.4 Trigonometric equations

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{{Mall:Vald flik|[[4.4 Trigonometriska ekvationer|Teori]]}}
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{{Selected tab|[[4.4 Trigonometric equations|Theory]]}}
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{{Not selected tab|[[4.4 Exercises|Exercises]]}}
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{{Info|
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'''Innehåll:'''
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'''Contents:'''
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*Trigonometriska grundekvationer
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* Simple trigonometric equations
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*Enklare trigonometriska ekvationer
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}}
}}
{{Info|
{{Info|
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'''Lärandemål:'''
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'''Learning outcomes: '''
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Efter detta avsnitt ska du ha lärt dig att:
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After this section, you will have learned how to:
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*Lösa trigonometriska grundekvationer.
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* Solve the basic equations of trigonometry.
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*Lösa trigonometriska ekvationer som kan återföras till ovanstående ekvationstyp.
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* Solve trigonometric equations that can be reduced to basic equations.
}}
}}
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== Grundekvationer ==
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== Basic equations ==
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Trigonometriska ekvationer kan vara mycket komplicerade, men det finns också många typer av trigonometriska ekvationer som man kan lösa med ganska enkla metoder. Här skall vi börja med att titta på de mest grundläggande trigonometriska ekvationerna, av typerna <math>\sin x = a</math>, <math>\cos x = a</math> och <math>\tan x = a</math>.
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Trigonometric equations can be very complicated, but there are also many types which can be solved using relatively simple methods. Here we shall start by looking at the most basic trigonometric equations, of the type <math>\sin x = a</math>, <math>\cos x = a</math> and <math>\tan x = a</math>.
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Dessa ekvationer har oändligt många lösningar, såvida inte omständigheterna begränsar antalet möjliga lösningar (t ex att man söker en spetsig vinkel).
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These equations usually have an infinite number of solutions, unless the circumstances limit the number of possible solutions (for example, if one is looking for an acute angle).
<div class="exempel">
<div class="exempel">
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'''Exempel 1'''
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''' Example 1'''
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Lös ekvationen <math>\,\sin x = \frac{1}{2}</math>.
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Solve the equation <math>\,\sin x = \frac{1}{2}</math>.
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Vår uppgift är att bestämma alla vinklar som gör att sinus av vinkeln blir <math>\tfrac{1}{2}</math>. Vi tar hjälp av enhetscirkeln. Notera att vinkeln här kallas <math>x</math>.
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Our task is to determine all the angles that have a sine equal to <math>\tfrac{1}{2}</math>. The unit circle helps us in this. Note that here the angle is designated as <math>x</math>.
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<center>{{:4.4 - Figur - Två enhetscirklar med vinklar π/6 resp. 5π/6}}</center>
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<center>{{:4.4 - Figure - Two unit circles with angles π/6 and 5π/6, respectively}}</center>
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I figuren har vi angivit de två riktningar som ger punkter med ''y''-koordinat <math>\tfrac{1}{2}</math> i enhetscirkeln, dvs. vinklar med sinusvärdet <math>\tfrac{1}{2}</math>. Den första är standardvinkeln <math>30^\circ = \pi / 6</math> och av symmetriskäl bildar den andra vinkeln <math>30^\circ</math> mot den negativa ''x''-axeln, vilket gör att den vinkeln är <math>180^\circ – 30^\circ = 150^\circ</math> eller i radianer <math>\pi – \pi / 6 = 5\pi / 6</math>. Detta är de enda lösningar till ekvationen <math>\sin x = \tfrac{1}{2}</math> mellan <math>0</math> och <math>2\pi</math>.
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The figure illustrates the two points on the circle which have ''y''-coordinate <math>\tfrac{1}{2}</math>, i.e. the two points whose corresponding angles have sine value <math>\tfrac{1}{2}</math>. The first is the standard angle <math>30^\circ = \pi / 6</math> and by symmetry the other angle makes <math>30^\circ</math> with the negative ''x''-axis. Therefore the other angle is <math>180^\circ – 30^\circ = 150^\circ</math> or in radians <math>\pi – \pi / 6 = 5\pi / 6</math>. These are the only solutions to the equation <math>\sin x = \tfrac{1}{2}</math> between <math>0</math> and <math>2\pi</math>.
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Vi kan dock lägga till ett godtyckligt antal varv till dessa två vinklar och fortfarande få samma sinusvärde. Alla vinklar med sinusvärde <math>\tfrac{1}{2}</math> är alltså
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However, we can add an arbitrary number of revolutions to these two angles and still get the same value for the sine . Thus all angles <math>x</math> for which <math>\sin x = \tfrac{1}{2}</math> are
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{{Fristående formel||<math>\begin{cases}
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{{Displayed math||<math>\begin{cases}
x &= \dfrac{\pi}{6} + 2n\pi\\
x &= \dfrac{\pi}{6} + 2n\pi\\
x &= \dfrac{5\pi}{6} + 2n\pi
x &= \dfrac{5\pi}{6} + 2n\pi
\end{cases}</math>}}
\end{cases}</math>}}
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där <math>n</math> är ett godtyckligt heltal. Detta kallas för den fullständiga lösningen till ekvationen.
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where <math>n</math> is an arbitrary integer. This is called the general solution to the equation.
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Lösningarna syns också i figuren nedan där grafen till <math>y = \sin x</math> skär linjen <math>y=\tfrac{1}{2}</math>.
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The solutions can also be obtained from the figure below, by looking at where the graph of <math>y = \sin x</math> intersects the line <math>y=\tfrac{1}{2}</math>.
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<center>{{:4.4 - Figur - Kurvorna y = sin x och y = ½}}</center>
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<center>{{:4.4 - Figure - The curves y = sin x and y = ½}}</center>
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 2'''
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''' Example 2'''
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Lös ekvationen <math>\,\cos x = \frac{1}{2}</math>.
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Solve the equation <math>\,\cos x = \frac{1}{2}</math>.
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Vi tar återigen hjälp av enhetscirkeln.
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We once again study the unit circle.
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<center>{{:4.4 - Figur - Två enhetscirklar med vinklar π/3 resp. -π/3}}</center>
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<center>{{:4.4 - Figure - Two unit circles with angles π/3 and -π/3, respectively}}</center>
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Vi vet att cosinus blir <math>\tfrac{1}{2}</math> för vinkeln <math>\pi/3</math>. Den enda andra riktning i enhetscirkeln som ger samma värde på cosinus har vinkeln <math>-\pi/3</math>. Lägger vi till ett helt antal varv till dessa vinklar får vi den fullständiga lösningen
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We know that cosine is <math>\tfrac{1}{2}</math> for the angle <math>\pi/3</math>. The only other point on the unit circle which has ''x''-coordinate <math>\frac{1}{2}</math> corresponds to the angle <math>-\pi/3</math>. Adding an integral number of revolutions to these angles we get the general solution
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{{Fristående formel||<math>x = \pm \pi/3 + n \cdot 2\pi\,\mbox{,}</math>}}
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{{Displayed math||<math>x = \pm \pi/3 + n \cdot 2\pi\,\mbox{,}</math>}}
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där <math>n</math> är ett godtyckligt heltal.
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where <math>n</math> is an arbitrary integer.
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 3'''
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''' Example 3'''
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Lös ekvationen <math>\,\tan x = \sqrt{3}</math>.
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Solve the equation <math>\,\tan x = \sqrt{3}</math>.
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En lösning till ekvationen är standardvinkeln <math>x=\pi/3</math>.
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One solution to the equation is the standard angle <math>x=\pi/3</math>.
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Om vi betraktar enhetscirkeln så är tangens av en vinkel lika med riktningskoefficienten för den räta linje genom origo som bildar vinkeln <math>x</math> med den positiva ''x''-axeln.
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If we study the unit circle then we see that the tangent of an angle is equal to the slope of the straight line through the origin which makes an angle <math>x</math> with the positive ''x''-axis .
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<center>{{:4.4 - Figur - Två enhetscirklar med vinklar π/3 resp. π+π/3}}</center>
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<center>{{:4.4 - Figure - Two unit circles with angles π/3 and π+π/3, respectively}}</center>
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Därför ser vi att lösningarna till <math>\tan x = \sqrt{3}</math> upprepar sig varje halvt varv <math>\pi/3</math>, <math>\pi/3 +\pi</math>, <math>\pi/3+ \pi +\pi</math> osv. Den fullständiga lösningen kan vi därmed få fram genom att utgå från lösningen <math>\pi/3</math> och lägga till eller dra ifrån multiplar av <math>\pi</math>,
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Therefore, we see that the solutions to <math>\tan x = \sqrt{3}</math> repeat themselves every half revolution, and so the general solution can be obtained from the solution <math>\pi/3</math> by adding or subtracting multiples of <math>\pi</math>:
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{{Fristående formel||<math>x = \pi/3 + n \cdot \pi\,\mbox{,}</math>}}
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{{Displayed math||<math>x = \pi/3 + n \cdot \pi\,\mbox{,}</math>}}
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där <math>n</math> är ett godtyckligt heltal.
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where <math>n</math> is an arbitrary integer.
</div>
</div>
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== Några mer komplicerade ekvationer ==
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== Somewhat more complicated equations ==
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Trigonometriska ekvationer kan se ut på många olika sätt, och det är omöjligt att här ge en fullständig genomgång av alla tänkbara ekvationer. Men låt oss studera några exempel, där vi kan ha nytta av att vi kan lösa grundekvationerna.
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Trigonometric equations can vary in many ways, and it is impossible to give a full catalogue of all possible equations. But let us study some examples where we can use our knowledge of solving basic equations.
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Vissa trigonometriska ekvationer kan förenklas genom att de skrivs om med hjälp av trigonometriska samband. Detta kan t ex leda till en andragradsekvation, som i nedanstående exempel där man använder att <math>\cos 2x = 2 \cos^2\!x – 1</math>.
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Some trigonometric equations can be simplified by rewriting them with the help of the trigonometric relationships. This, for example, could lead to a quadratic equation, as in the example below.
<div class="exempel">
<div class="exempel">
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'''Exempel 4'''
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''' Example 4'''
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Lös ekvationen <math>\,\cos 2x – 4\cos x + 3= 0</math>.
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Solve the equation <math>\,\cos 2x – 4\cos x + 3= 0</math>.
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Omskrivning med hjälp av formeln <math>\cos 2x = 2 \cos^2\!x – 1</math> ger
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Rewrite by using the formula <math>\cos 2x = 2 \cos^2\!x – 1</math>, so that
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{{Fristående formel||<math>(2 \cos^2\!x – 1) – 4\cos x + 3 = 0\,\mbox{,}</math>}}
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{{Displayed math||<math>(2 \cos^2\!x – 1) – 4\cos x + 3 = 0\,\mbox{.}</math>}}
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vilket kan förenklas till ekvationen (efter division med 2)
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Simplifying, we see that
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{{Fristående formel||<math>\cos^2\!x - 2 \cos x +1 =0\,\mbox{.}</math>}}
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{{Displayed math||<math>\cos^2\!x - 2 \cos x +1 =0\,\mbox{.}</math>}}
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Vänsterledet kan faktoriseras med kvadreringsregeln till
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The left-hand side can factorised by using the squaring rule to give
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{{Fristående formel||<math>(\cos x-1)^2 = 0\,\mbox{.}</math>}}
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{{Displayed math||<math>(\cos x-1)^2 = 0\,\mbox{.}</math>}}
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Denna ekvation kan bara vara uppfylld om <math>\cos x = 1</math>. Grundekvationen <math>\cos x=1</math> kan vi lösa på det vanliga sättet och den fullständiga lösningen är
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This equation can only be satisfied if <math>\cos x = 1</math>. The basic equation <math>\cos x=1</math> can be solved in the normal way, and thus the complete solution is
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{{Fristående formel||<math>
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{{Displayed math||<math>
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x = 2n\pi \qquad (\,n \mbox{ godtyckligt heltal).}</math>}}
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x = 2n\pi \qquad (\,n \mbox{ arbitrary integer).}</math>}}
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 5'''
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''' Example 5'''
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Lös ekvationen <math>\,\frac{1}{2}\sin x + 1 – \cos^2 x = 0</math>.
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Solve the equation <math>\,\frac{1}{2}\sin x + 1 – \cos^2 x = 0</math>.
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Enligt den trigonometriska ettan är <math>\sin^2\!x + \cos^2\!x = 1</math>, dvs. <math>1 – \cos^2\!x = \sin^2\!x</math>.
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According to the Pythagorean identity <math>\sin^2\!x + \cos^2\!x = 1</math>, i.e. <math>1 – \cos^2\!x = \sin^2\!x</math>. Thus the equation can be written as
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Ekvationen kan alltså skrivas
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{{Displayed math||<math>\tfrac{1}{2}\sin x + \sin^2\!x = 0\,\mbox{.}</math>}}
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{{Fristående formel||<math>\tfrac{1}{2}\sin x + \sin^2\!x = 0\,\mbox{.}</math>}}
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Genom att nu bryta ut en faktor <math>\sin x</math> får vi
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Factorising out <math>\sin x</math> we get
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{{Fristående formel||<math>
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{{Displayed math||<math>
\sin x\,\cdot\,\bigl(\tfrac{1}{2} + \sin x\bigr) = 0 \, \mbox{.}</math>}}
\sin x\,\cdot\,\bigl(\tfrac{1}{2} + \sin x\bigr) = 0 \, \mbox{.}</math>}}
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Från denna faktoriserade form av ekvationen ser vi att lösningarna antingen måste uppfylla <math>\sin x = 0</math> eller <math>\sin x = -\tfrac{1}{2}</math>, vilka är två vanliga grundekvationer på formen <math>\sin x = a</math> och kan lösas som i exempel 1. Lösningarna blir till slut
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From this factorised form of the equation, we see that the solutions either have to satisfy <math>\sin x = 0</math> or <math>\sin x = -\tfrac{1}{2}</math>, which are two basic equations of the type <math>\sin x = a</math> and can be solved as in Example 1. The solutions turn out to be
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{{Fristående formel||<math>
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{{Displayed math||<math>
\begin{cases}
\begin{cases}
x &= n\pi\\
x &= n\pi\\
Line 138: Line 136:
x &= 7\pi/6+2n\pi
x &= 7\pi/6+2n\pi
\end{cases}
\end{cases}
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\qquad (\,n\ \text{godtyckligt heltal})\mbox{.}</math>}}
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\qquad (\,n\ \text{ arbitrary integer})\mbox{.}</math>}}
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 6'''
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''' Example 6'''
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Lös ekvationen <math>\,\sin 2x =4 \cos x</math>.
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Solve the equation <math>\,\sin 2x =4 \cos x</math>.
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Genom omskrivning med formeln för dubbla vinkeln blir ekvationen
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By rewriting the equation using the formula for double-angles we get
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{{Fristående formel||<math>2\sin x\,\cos x – 4 \cos x = 0\,\mbox{.}</math>}}
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{{Displayed math||<math>2\sin x\,\cos x – 4 \cos x = 0\,\mbox{.}</math>}}
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Vi delar båda led med 2 och bryter ut en faktor <math>\cos x</math>, vilket ger
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Dividing both sides by 2 and factorising out <math>\cos x</math>, gives
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{{Fristående formel||<math>\cos x\,\cdot\,( \sin x – 2) = 0\,\mbox{.}</math>}}
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{{Displayed math||<math>\cos x\,\cdot\,( \sin x – 2) = 0\,\mbox{.}</math>}}
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Eftersom produkten bara kan bli noll genom att en faktor är noll, så kan ekvationen delas upp i grundekvationerna
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The left-hand side can only be zero if one of the factors is zero, and we have reduced the original equation into two basic equations:
* <math>\cos x = 0</math>,
* <math>\cos x = 0</math>,
* <math>\sin x = 2</math>.
* <math>\sin x = 2</math>.
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Men <math>\sin x</math> kan aldrig bli större än 1, så ekvationen <math>\sin x = 2</math> saknar lösningar. Då återstår bara
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However, <math>\sin x</math> can never be greater than 1, so the equation <math>\sin x = 2</math> has no solutions. That just leaves <math>\cos x = 0</math>, and using the unit circle we see that the general solution is
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<math>\cos x = 0</math>, vilken med hjälp av enhetscirkeln ger den fullständiga lösningen <math>x = \pi / 2 + n \cdot \pi</math>.
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{{Displayed math||<math>x = \frac{\pi}{2} + n\pi\mbox{,}</math>}}
 +
for <math>n</math> an arbitrary integer.
 +
 
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 7'''
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''' Example 7'''
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Lös ekvationen <math>\,4\sin^2\!x – 4\cos x = 1</math>.
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Solve the equation <math>\,4\sin^2\!x – 4\cos x = 1</math>.
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Med den trigonometriska ettan kan <math>\sin^2\!x</math> bytas ut mot <math>1 – \cos^2\!x</math>. Då får vi
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Using the Pythagorean identity we can replace <math>\sin^2\!x</math> by <math>1 – \cos^2\!x</math>. Then{{Displayed math||<math>
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{{Fristående formel||<math>
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\begin{align*}
\begin{align*}
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4 (1 – \cos^2\!x) – 4 \cos x &= 1\,\mbox{,}\\
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4 (1 – \cos^2\!x)\ –\ 4 \cos x &= 1\,\\
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4 – 4 \cos^2\!x – 4 \cos x &= 1\,\mbox{,}\\
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\Rightarrow 4\ –\ 4 \cos^2\!x\ –\ 4 \cos x &= 1\,\\
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–4\cos^2\!x – 4 \cos x + 4 – 1 &= 0\,\mbox{,}\\
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\Rightarrow –4\cos^2\!x\ –\ 4 \cos x\ +\ 4\ –\ 1 &= 0\,\\
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\cos^2\!x + \cos x – \tfrac{3}{4} &= 0\,\mbox{.}\\
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\Rightarrow \cos^2\!x + \cos x – \tfrac{3}{4} &= 0\,\mbox{.}\\
\end{align*}</math>}}
\end{align*}</math>}}
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Detta är en andragradsekvation i <math>\cos x</math>, som har lösningarna
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This is a quadratic equation in <math>\cos x</math>, which has the solutions
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{{Fristående formel||<math>
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{{Displayed math||<math>
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\cos x = -\tfrac{3}{2} \quad\text{och}\quad
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\cos x = -\tfrac{3}{2} \quad\text{and}\quad
\cos x = \tfrac{1}{2}\,\mbox{.}</math>}}
\cos x = \tfrac{1}{2}\,\mbox{.}</math>}}
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Eftersom värdet av <math>\cos x</math> ligger mellan <math>–1</math> och <math>1</math> kan ekvationen <math>\cos x=-\tfrac{3}{2}</math> inte ha några lösningar. Då återstår bara grundekvationen
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Since the value of <math>\cos x</math> is between <math>–1</math> and <math>1</math>, the equation <math>\cos x=-\tfrac{3}{2}</math> has no solutions. That leaves only the basic equation
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{{Fristående formel||<math>\cos x = \tfrac{1}{2}\,\mbox{,}</math>}}
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{{Displayed math||<math>\cos x = \tfrac{1}{2}\,\mbox{,}</math>}}
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som löses enligt exempel 2.
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which may be solved as in Example 2.
</div>
</div>
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[[4.4 Övningar|Övningar]]
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[[4.4 Exercises|Exercises]]
<div class="inforuta" style="width:580px;">
<div class="inforuta" style="width:580px;">
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'''Råd för inläsning'''
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'''Study advice'''
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'''Grund- och slutprov'''
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Efter att du har läst texten och arbetat med övningarna ska du göra grund- och slutprovet för att bli godkänd på detta avsnitt. Du hittar länken till proven i din student lounge.
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'''Tänk på att:'''
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Det är bra om man lär sig de vanliga trigonometriska formlerna (identiteterna) och övar upp en viss vana på att förenkla och manipulera trigonometriska uttryck.
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'''Basic and final tests'''
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Det är viktigt att man lär sig de grundläggande ekvationerna, av typen <math>\sin x = a</math>, <math>\cos x = a</math> eller <math>\tan x = a</math> (där <math>a</math> är ett reellt tal). Det är också viktigt att man vet att dessa ekvationer har oändligt många lösningar.
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After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.
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'''Lästips'''
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'''Keep in mind that...'''
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för dig som vill fördjupa dig ytterligare eller behöver en längre förklaring vill vi tipsa om:
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It is a good idea to learn the most common trigonometric formulas (identities) and practice simplifying and manipulating trigonometric expressions.
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[http://www.theducation.se/kurser/umaprep/4_trigonometri/44_trig_ekvationer/index.asp Läs mer om trigonometriska ekvationer i Theducations gymnasielexikon ]
 
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[http://www.theducation.se/kurser/umaprep/4_trigonometri/44_trig_ekvationer/445_typ_asinx/index.asp Träna på trigonometriska räkneexempel i Theducations gymnasielexikon]
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It is important to be familiar with the basic equations, such as <math>\sin x = a</math>, <math>\cos x = a</math> or <math>\tan x = a</math> (where <math>a</math> is a real number). It is also important to know that these equations typically have infinitely many solutions.
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'''Länktips'''
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'''Useful web sites'''
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[http://www.ies.co.jp/math/java/trig/ABCsinX/ABCsinX.html Experimentera med grafen y=a sin b(x-c)]
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[http://www.ies.co.jp/math/java/trig/ABCsinX/ABCsinX.html Experiment with the graph y = a sin b (x-c) ]
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[http://www.theducation.se/kurser/experiment/gyma/applets/ex45_derivatasinus/Ex45Applet.html Experimentera med derivatan av sin x]
 
</div>
</div>

Current revision

       Theory          Exercises      

Contents:

  • Simple trigonometric equations

Learning outcomes:

After this section, you will have learned how to:

  • Solve the basic equations of trigonometry.
  • Solve trigonometric equations that can be reduced to basic equations.

Basic equations

Trigonometric equations can be very complicated, but there are also many types which can be solved using relatively simple methods. Here we shall start by looking at the most basic trigonometric equations, of the type \displaystyle \sin x = a, \displaystyle \cos x = a and \displaystyle \tan x = a.

These equations usually have an infinite number of solutions, unless the circumstances limit the number of possible solutions (for example, if one is looking for an acute angle).

Example 1

Solve the equation \displaystyle \,\sin x = \frac{1}{2}.


Our task is to determine all the angles that have a sine equal to \displaystyle \tfrac{1}{2}. The unit circle helps us in this. Note that here the angle is designated as \displaystyle x.

[Image]

The figure illustrates the two points on the circle which have y-coordinate \displaystyle \tfrac{1}{2}, i.e. the two points whose corresponding angles have sine value \displaystyle \tfrac{1}{2}. The first is the standard angle \displaystyle 30^\circ = \pi / 6 and by symmetry the other angle makes \displaystyle 30^\circ with the negative x-axis. Therefore the other angle is \displaystyle 180^\circ – 30^\circ = 150^\circ or in radians \displaystyle \pi – \pi / 6 = 5\pi / 6. These are the only solutions to the equation \displaystyle \sin x = \tfrac{1}{2} between \displaystyle 0 and \displaystyle 2\pi.

However, we can add an arbitrary number of revolutions to these two angles and still get the same value for the sine . Thus all angles \displaystyle x for which \displaystyle \sin x = \tfrac{1}{2} are

\displaystyle \begin{cases} 
   x &= \dfrac{\pi}{6} + 2n\pi\\
   x &= \dfrac{5\pi}{6} + 2n\pi
 \end{cases}

where \displaystyle n is an arbitrary integer. This is called the general solution to the equation.

The solutions can also be obtained from the figure below, by looking at where the graph of \displaystyle y = \sin x intersects the line \displaystyle y=\tfrac{1}{2}.

[Image]

Example 2

Solve the equation \displaystyle \,\cos x = \frac{1}{2}.


We once again study the unit circle.

[Image]

We know that cosine is \displaystyle \tfrac{1}{2} for the angle \displaystyle \pi/3. The only other point on the unit circle which has x-coordinate \displaystyle \frac{1}{2} corresponds to the angle \displaystyle -\pi/3. Adding an integral number of revolutions to these angles we get the general solution

\displaystyle x = \pm \pi/3 + n \cdot 2\pi\,\mbox{,}

where \displaystyle n is an arbitrary integer.

Example 3

Solve the equation \displaystyle \,\tan x = \sqrt{3}.


One solution to the equation is the standard angle \displaystyle x=\pi/3.

If we study the unit circle then we see that the tangent of an angle is equal to the slope of the straight line through the origin which makes an angle \displaystyle x with the positive x-axis .

[Image]

Therefore, we see that the solutions to \displaystyle \tan x = \sqrt{3} repeat themselves every half revolution, and so the general solution can be obtained from the solution \displaystyle \pi/3 by adding or subtracting multiples of \displaystyle \pi:

\displaystyle x = \pi/3 + n \cdot \pi\,\mbox{,}

where \displaystyle n is an arbitrary integer.


Somewhat more complicated equations

Trigonometric equations can vary in many ways, and it is impossible to give a full catalogue of all possible equations. But let us study some examples where we can use our knowledge of solving basic equations.

Some trigonometric equations can be simplified by rewriting them with the help of the trigonometric relationships. This, for example, could lead to a quadratic equation, as in the example below.

Example 4

Solve the equation \displaystyle \,\cos 2x – 4\cos x + 3= 0.


Rewrite by using the formula \displaystyle \cos 2x = 2 \cos^2\!x – 1, so that

\displaystyle (2 \cos^2\!x – 1) – 4\cos x + 3 = 0\,\mbox{.}

Simplifying, we see that

\displaystyle \cos^2\!x - 2 \cos x +1 =0\,\mbox{.}

The left-hand side can factorised by using the squaring rule to give

\displaystyle (\cos x-1)^2 = 0\,\mbox{.}

This equation can only be satisfied if \displaystyle \cos x = 1. The basic equation \displaystyle \cos x=1 can be solved in the normal way, and thus the complete solution is

\displaystyle 
 x = 2n\pi \qquad (\,n \mbox{ arbitrary integer).}

Example 5

Solve the equation \displaystyle \,\frac{1}{2}\sin x + 1 – \cos^2 x = 0.


According to the Pythagorean identity \displaystyle \sin^2\!x + \cos^2\!x = 1, i.e. \displaystyle 1 – \cos^2\!x = \sin^2\!x. Thus the equation can be written as

\displaystyle \tfrac{1}{2}\sin x + \sin^2\!x = 0\,\mbox{.}

Factorising out \displaystyle \sin x we get

\displaystyle 
 \sin x\,\cdot\,\bigl(\tfrac{1}{2} + \sin x\bigr) = 0 \, \mbox{.}

From this factorised form of the equation, we see that the solutions either have to satisfy \displaystyle \sin x = 0 or \displaystyle \sin x = -\tfrac{1}{2}, which are two basic equations of the type \displaystyle \sin x = a and can be solved as in Example 1. The solutions turn out to be

\displaystyle 
 \begin{cases}
   x &= n\pi\\
   x &= -\pi/6+2n\pi\\
   x &= 7\pi/6+2n\pi
 \end{cases}
 \qquad (\,n\ \text{ arbitrary integer})\mbox{.}

Example 6

Solve the equation \displaystyle \,\sin 2x =4 \cos x.


By rewriting the equation using the formula for double-angles we get

\displaystyle 2\sin x\,\cos x – 4 \cos x = 0\,\mbox{.}

Dividing both sides by 2 and factorising out \displaystyle \cos x, gives

\displaystyle \cos x\,\cdot\,( \sin x – 2) = 0\,\mbox{.}

The left-hand side can only be zero if one of the factors is zero, and we have reduced the original equation into two basic equations:

  • \displaystyle \cos x = 0,
  • \displaystyle \sin x = 2.

However, \displaystyle \sin x can never be greater than 1, so the equation \displaystyle \sin x = 2 has no solutions. That just leaves \displaystyle \cos x = 0, and using the unit circle we see that the general solution is

\displaystyle x = \frac{\pi}{2} + n\pi\mbox{,}

for \displaystyle n an arbitrary integer.

Example 7

Solve the equation \displaystyle \,4\sin^2\!x – 4\cos x = 1.


Using the Pythagorean identity we can replace \displaystyle \sin^2\!x by \displaystyle 1 – \cos^2\!x. Then

\displaystyle 
 \begin{align*}
   4 (1 – \cos^2\!x)\ –\ 4 \cos x &= 1\,\\
   \Rightarrow 4\ –\ 4 \cos^2\!x\ –\ 4 \cos x &= 1\,\\
   \Rightarrow –4\cos^2\!x\ –\ 4 \cos x\ +\ 4\ –\ 1 &= 0\,\\
   \Rightarrow \cos^2\!x + \cos x – \tfrac{3}{4} &= 0\,\mbox{.}\\
 \end{align*}

This is a quadratic equation in \displaystyle \cos x, which has the solutions

\displaystyle 
 \cos x = -\tfrac{3}{2} \quad\text{and}\quad
 \cos x = \tfrac{1}{2}\,\mbox{.}

Since the value of \displaystyle \cos x is between \displaystyle –1 and \displaystyle 1, the equation \displaystyle \cos x=-\tfrac{3}{2} has no solutions. That leaves only the basic equation

\displaystyle \cos x = \tfrac{1}{2}\,\mbox{,}

which may be solved as in Example 2.


Exercises

Study advice

Basic and final tests

After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.


Keep in mind that...

It is a good idea to learn the most common trigonometric formulas (identities) and practice simplifying and manipulating trigonometric expressions.


It is important to be familiar with the basic equations, such as \displaystyle \sin x = a, \displaystyle \cos x = a or \displaystyle \tan x = a (where \displaystyle a is a real number). It is also important to know that these equations typically have infinitely many solutions.


Useful web sites

Experiment with the graph y = a sin b (x-c)

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