Trigonometric equations

Summary and examples

Trigonometric equations are equations that contain one or more trigonometric functions. We will restrict ourselves to a few common trigonometric equations.
In solving such equations a (graphical) calculator is often - but not always -necessary. We try to limit the use of such calculators to a minimum, which increases understanding.
For example, let us look at the equation:

$\sin(x)=\displaystyle\frac{1}{2}$

Suppose we want to solve this equation with a calculator. We can find the $x$ as follows:

$\sin^{-1}(\displaystyle\frac{1}{2})=0.524$

Here $\sin^{-1}$ is the inverse function of the function $\sin$ . The latter function gives the value $\sin(x)$ for a certain value of $x$ , the inverse function does the inverse. If we give the value of $\sin(x)$ we get the value of $x$ in return. We can check this: when we want to calculate the inverse $sin$ function of the value $0.524$ (i.e. $\pi/6$ ) we get $0.5$ (not exactly because $0.524$ is an approximation of $\pi/6$ ).

We could have solved this equation without a calculator. The table with trigonometric values shows that $\frac{1}{2}$ corresponds to the value:

$x=\displaystyle\frac{\pi}{6}$

We are not finished yet. Looking at the unit circle (see also Unit circle and simple formulas), we find that within the circle there is another value of $x$ for which the $sin$ function gives the value $0.5$ . The first value lies in the first quadrant, but in the second quadrant there is another solution, namely:

$x=\pi-\displaystyle\frac{\pi}{6}=\displaystyle\frac{5}{6}\pi$

This value of $x$ is a solution of the equation as well. Actually, we know that also the following values of $x$ are solutions:

$x=\displaystyle\frac{\pi}{6}+2k\pi, k=0,\pm1, \pm2, ...$

because the period of the $sin$ -function is $2\pi$ .

Thus we have:

$x=\displaystyle\frac{5}{6}\pi+2k\pi, k=0, \pm1, \pm2, ...$

In general the following holds.

If:

$x=q$

is the solution of the equation that is found using a table or calculator:

$\sin(x)=p$

then the solutions of this equation are:

$x=q+2k\pi, k=0, \pm1, \pm2, ...$

or:

$x=\pi-q+2k\pi, k=0, \pm1, \pm2, ...$

In a similar way we solve the equation:

$\cos(x)=p$

If:

$x=q$

is the solution of the equation that is found using a table or calculator:

$\cos(x)=p$

then the solutions of this equation are:

$x=q+2k\pi, k=0, \pm1, \pm2, ...$

or:

$x=-q+2k\pi, k=0, \pm1, \pm2, ...$

Let us look at equations having the form:

$\sin(x)=\sin(y)$

or:

$\cos(x)=\cos(y)$

These equations are solved in a similar way as shown above. The first type of these equations have the following solutions:

$x=y+2k\pi, k=0, \pm1, \pm2, ...$

or:

$x=\pi-y+2k\pi, k=0,\pm1, \pm2, ...$

For the second type of the equations we have:

$x=y+2k\pi, k=0, \pm1, \pm2, ...$

or:

$x=-y+2k\pi, k=0, \pm1, \pm2, ...$

There are also equations of the type:

$\sin(x)=\cos(y)$

In this case we may use the following formulas which can be deduced straightforward from the unit circle (see also Unit circle and simple formulas):

$\sin(x)=\cos(\displaystyle\frac{\pi}{2}-x)$

$\cos(x)=\sin(\displaystyle\frac{\pi}{2}-x)$

Example 1

Solve:

$\sin(x)=\displaystyle\frac{1}{2}\sqrt{2}$

The table shows:

$x=\displaystyle\frac{\pi}{4}$

Then we also know that a solution is:

$x=\pi-\displaystyle\frac{\pi}{4}=\displaystyle\frac{3}{4}\pi$

Because the $\sin$ -function has the period $2\pi$ , the solutions of the equation are:

$x=\displaystyle\frac{\pi}{4}+2k\pi, k=0, \pm1, \pm2, ...$

$x=\displaystyle\frac{3}{4}\pi+2k\pi, k=0, \pm1, \pm2, ...$

Example 2

Solve:

$\cos(x)=\displaystyle\frac{1}{2}$

The table shows:

$x=\displaystyle\frac{\pi}{3}$

Because the $\cos$ -function has the period $2\pi$ , the solutions of the equation are:

$x=\displaystyle\frac{\pi}{3}+2k\pi, k=0, \pm1, \pm2, ...$

or:

$x=-\displaystyle\frac{\pi}{3}+2k\pi, k=0, \pm1, \pm2, ...$

Example 3

Solve:

$\sin(x)=0.6$

In this case the value cannot be taken from a table and thus we need to use a calculator.

We find (approximated):

$x=0.64$

and thus (also approximated):

$x=3.14-0.64=2.50$

Because the $\sin$ -function has the period $2\pi$ , the solutions of the equation are (approximated):

$x=0.64+2k\pi, k=0, \pm1, \pm2, ...$

$x=2.50+2k\pi, k=0, \pm1, \pm2, ...$

Example 4

Solve:

$\sin(x)=1.2$

In this case table nor calculator can help us, because the $\sin$ -function lies in the interval $[-1,+1]$ (the so-called range) and therefore this equation does not have a solution.