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 equal to \displaystyle \tfrac{1}{2}. The unit circle helps us in this. Note that here the angle is designated as \displaystyle x.

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

   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}.

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 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

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 .

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:

where \displaystyle n is 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, so that

Simplifying, we see that

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 thus 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. Thus the equation can be written as

Factorising out \displaystyle \sin x we get

 \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{ 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

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

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

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

 \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

 \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

which may be solved as in Example 2.