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:


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


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:


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:


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.



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


then the solutions of this equation are:

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


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

In a similar way we solve the equation:




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


then the solutions of this equation are:

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


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

Let us look at equations having the form:




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, ...


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, ...


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

There are also equations of the type:


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):



Example 1 



The table shows:


Then we also know that a solution is:


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



The table shows:


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, ...


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

Example 3 



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

We find (approximated):


and thus (also approximated):


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



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.

Example 5



The solution is straightforward:

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

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

Web Design BangladeshWeb Design BangladeshMymensingh