4.4 Trigonometric equations

Example 1

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

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

In the figure, we have shown the two directions that give us points which have a y- coordinate \displaystyle \tfrac{1}{2} on the unit circle, i.e. angles with a 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, This means that the 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 with a value of the sine \displaystyle \tfrac{1}{2} are

   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 also seen in the figure below where the graph of \displaystyle y = \sin x intersect the line \displaystyle y=\tfrac{1}{2}.

Example 2

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

We once again study the unit circle.

We know that cosine is \displaystyle \tfrac{1}{2} for the angle \displaystyle \pi/3. The only other direction in the unit circle, which produces the same value for the cosine is the angle \displaystyle -\pi/3. Adding an integral number of revolutions to these angles we get the general solution

where \displaystyle n is an arbitrary integer.

Example 3

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

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

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

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

where \displaystyle n s an arbitrary integer.

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 giving

which can be simplified to the equation (after division by 2)

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

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 the complete solution is

 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. The equation can be written as

Factorising out \displaystyle \sin x one gets

 \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

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

Example 6

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

By rewriting the equation using the formula for double-angles one gets

We divide both sides with 2 and factorise out \displaystyle \cos x, which gives

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

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

But \displaystyle \sin x can never be greater than 1, so the equation \displaystyle \sin x = 2 has no solutions. That leaves just \displaystyle \cos x = 0, and using the unit circle gives the general solution \displaystyle x = \pi / 2 + n \cdot \pi.

Example 7

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

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

 \begin{align*}
   4 (1 – \cos^2\!x) – 4 \cos x &= 1\,\mbox{,}\\
   4 – 4 \cos^2\!x – 4 \cos x &= 1\,\mbox{,}\\
   –4\cos^2\!x – 4 \cos x + 4 – 1 &= 0\,\mbox{,}\\
   \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

 \cos x = -\tfrac{3}{2} \quad\text{och}\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

that may be solved as in example 2.