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Solution 4.4:7a

From Förberedande kurs i matematik 1

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If we examine the equation, we see that <math>x</math> only occurs as <math>\sin x</math> and it can therefore be appropriate to take an intermediary step and solve for <math>\sin x</math>, instead of trying to solve for <math>x</math> directly.
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<center> [[Bild:4_4_7a-1(2).gif]] </center>
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If we write <math>t = \sin x</math> and treat <math>t</math> as a new unknown variable, the equation becomes
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<center> [[Bild:4_4_7a-2(2).gif]] </center>
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{{Displayed math||<math>2t^2 + t = 1</math>}}
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and it is expressed completely in terms of <math>t</math>. This is a normal quadratic equation; after dividing by 2, we complete the square on the left-hand side,
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{{Displayed math||<math>\begin{align}
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t^2 + \frac{1}{2}t - \frac{1}{2}
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&= \Bigl(t+\frac{1}{4}\Bigr)^2 - \Bigl(\frac{1}{4}\Bigr)^2 - \frac{1}{2}\\[5pt]
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&= \Bigl(t+\frac{1}{4}\Bigr)^2 - \frac{9}{16}
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\end{align}</math>}}
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and then obtain the equation
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{{Displayed math||<math>\Bigl(t+\frac{1}{4}\Bigr)^2 = \frac{9}{16}</math>}}
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which has the solutions <math>t=-\tfrac{1}{4}\pm \sqrt{\tfrac{9}{16}}=-\tfrac{1}{4}\pm \tfrac{3}{4}</math>, i.e.
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{{Displayed math||<math>t = -\frac{1}{4}+\frac{3}{4} = \frac{1}{2}\qquad\text{and}\qquad t = -\frac{1}{4}-\frac{3}{4} = -1\,\textrm{.}</math>}}
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Because <math>t=\sin x</math>, this means that the values of ''x'' that satisfy the equation in the exercise will necessarily satisfy one of the basic equations <math>\sin x = \tfrac{1}{2}</math> or <math>\sin x = -1\,</math>.
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<math>\sin x = \frac{1}{2}</math>:
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This equation has the solutions <math>x = \pi/6</math> and <math>x = \pi - \pi/6 = 5\pi/6</math> in the unit circle and the general solution is
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{{Displayed math||<math>x = \frac{\pi}{6}+2n\pi\qquad\text{and}\qquad x = \frac{5\pi}{6}+2n\pi\,,</math>}}
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where ''n'' is an arbitrary integer.
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<math>\sin x = -1</math>:
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The equation has only one solution <math>x = 3\pi/2</math> in the unit circle, and the general solution is therefore
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{{Displayed math||<math>x = \frac{3\pi}{2} + 2n\pi\,,</math>}}
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where ''n'' is an arbitrary integer.
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All of the solution to the equation are given by
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{{Displayed math||<math>\left\{\begin{align}
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x &= \pi/6+2n\pi\,,\\[5pt]
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x &= 5\pi/6+2n\pi\,,\\[5pt]
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x &= 3\pi/2+2n\pi\,,
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\end{align}\right.</math>}}
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where ''n'' is an arbitrary integer.

Current revision

If we examine the equation, we see that x only occurs as sinx and it can therefore be appropriate to take an intermediary step and solve for sinx, instead of trying to solve for x directly.

If we write t=sinx and treat t as a new unknown variable, the equation becomes

2t2+t=1

and it is expressed completely in terms of t. This is a normal quadratic equation; after dividing by 2, we complete the square on the left-hand side,

t2+21t−21=t+412−412−21=t+412−916

and then obtain the equation

t+412=916 

which has the solutions t=−41916=−4143 , i.e.

t=−41+43=21andt=−41−43=−1.


Because t=sinx, this means that the values of x that satisfy the equation in the exercise will necessarily satisfy one of the basic equations sinx=21 or sinx=−1.


sinx=21:

This equation has the solutions x=6 and x=−6=56 in the unit circle and the general solution is

x=6+2nandx=65+2n

where n is an arbitrary integer.


sinx=−1:

The equation has only one solution x=32 in the unit circle, and the general solution is therefore

x=23+2n

where n is an arbitrary integer.


All of the solution to the equation are given by

xxx=6+2n=56+2n=32+2n

where n is an arbitrary integer.

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