Trigonometric equations can be solved by first manipulating them into the form

$\sin(x)=a$,   $\cos(y)=b$   or   $\tan(z)=c$.

This could be very simple, a straight forward manipulation of a linear equation (Examples 1 and 2) or it could be a more complicated manipulation of a quadratic equation (Examples 3 and 4). Once in this format, one can use the $\sin^{-1}$, $\cos^{-1}$ and $\tan^{-1}$ buttons on a scientific calculator. However, at this stage it is very important to capture all solutions of the trigonometric equation since the scientific calculator will only give one default solution.

## Solving trigonometric equations using the trigonometric graphs

Once in the form described above, a calculator will only give you a default solution, it is up to you to find the rest. Remind yourselves of the trigonometric graphs. The following rules can be used to find all solutions on a given interval:

$y=\cos(x)$

The local lines of symmetry on the graph of cos(x) are every 180 degrees starting at 0. If there is a solution at 60, for example, there is also one at 360-60=300. Both of these solutions then repeat every 360 degrees.

$y=\sin(x)$

The local lines of symmetry on the graph of sin(x) are every 180 degrees starting at 90. If there is a solution at 30, for example, there is also one at 180-30=150. Both of these solutions then repeat every 360 degrees.

$y=\tan(x)$

The solutions on the tan graph are easier to find, the default value simply repeats every 180 degrees. If there is a solution at 45 degrees, for example, there is another at -135, 225, etc.

The tan graph is different to the sin and cos graphs. Firstly, tan can take a value between -infinity and infinity whereas sin and cos take values between -1 and 1. The graph of tan also has asymptotes as a result of dividing by cos(x)=0 in the expression tan(x)=sin(x)/cos(x).

It may be that the argument of the trigonometric function is a function of x. In these cases it is very important to extend the interval of solution values so that the full set of solutions are obtained. See Examples 2 and 4. Some students prefer to use ‘CAST’ – this is a complicated method for finding all of the solutions of trigonometric equations. See ‘CAST’ on YouTube. It is considerably simpler, as you will see in the examples below, to extend the interval of solutions according to the operations performed on x.

## Linear Trigonometric Equations

### Example 1

1. Solve $\sqrt{2}\sin(x)=\cos(x)$ on the interval $-360^\circ\leq x\leq 360^\circ$.
2. Solve $\tan(x+75^\circ)=3$ on the interval $0^\circ\leq x\leq 360^\circ$.

1. The equation is rearranged to give $\frac{\sin(x)}{\cos(x)}\equiv\tan(x)=\frac{1}{\sqrt{2}}\approx 0.707$. Inverting tan gives the default solution of $x=\tan^{-1}(0.707)=35.26^\circ$ and we can see from the graph that solutions repeat every 180 degrees:The solutions are $x=-324.74^\circ, -144.74^\circ, 35.26^\circ, 215.26^\circ$.
2. Similarly to the above, the default solution of the equation is $x+75^\circ=\tan^{-1}(3)=71.57^\circ$. There will be other solutions but we need to assess the intervals in which they will lie. Let $\theta=x+75^\circ$. If $0^\circ\leq x\leq 360^\circ$ then $75^\circ\leq \theta \leq 435^\circ$. The default value is outside this range and the solutions on the tan graph that lie in this interval are $\theta=251.57^\circ, 431.57^\circ$. It follows that the solutions to the original equation are $x=176.57^\circ$ and $x=356.57^\circ$.

### Example 2

Solve the following for $-180^\circ\leq x\leq 180^\circ$:

$4\cos(2x)-5=-1$

It is very important to note here that if x can be between $-180^\circ$ and $180^\circ$ then 2x can be between $-360^\circ$ and $360^\circ$. Firstly, as suggested above, we write the equation in the form $\cos(2x)=1$. Rather than transform the graph of cos(x), let $\theta=2x$ and look at the graph of $\cos(\theta)$. When looking at the graph of cos(theta) to find values where cos takes the value 1, it is important to check all values of theta between $-360^\circ$ and $360^\circ$.

On this interval, $\cos(\theta)=1$ when $\theta=2x$ is equal to $-360^\circ$, $0^\circ$ and $360^\circ$. This gives three solutions for x on the interval $-180^\circ\leq x\leq 180^\circ$:

$x=-180^\circ$, $x=0^\circ$ and $x=180^\circ$.

It is very common for students to lose marks on questions like this – they forget to extend the interval that caters for all possible solutions.

### Example 3

Solve the following for $0\leq x\leq 360^\circ$:

$\sin^2(x)=\sin(x)$

Don’t begin by dividing by sin(x) – this is because if you divide by anything you assume that it isn’t 0 and, of course, it might be. Instead, subtract sin(x) from both sides and factorise:

$\sin(x)(\sin(x)-1)=0$

A product that results in 0 means that either factor can be 0 and so we have sin(x)=0 or sin(x)-1=0, i.e. sin(x)=1.

By examining the graph of sin(x) (see more on trig graphs) and using the inverse sine button on your calculator to find the multiple solutions, we can see that the solutions on the interval $0\leq x\leq 360^\circ$ are $0^\circ, 90^\circ, 180^\circ,$ and $360^\circ$.

### Example 4

Solve

$6\sin^2(3\theta)=4-\cos(3\theta)$

on the interval $0\textless\theta\textless180^\circ$.

Firstly, recognise that this is almost a quadratic. Replacing $\sin^2(3\theta)$ with $1-\cos^2(3\theta)$ and rearranging, we have a quadratic in cos:

$6(1-\cos^2(3\theta))=4-\cos(3\theta)$

$\Longrightarrow 6\cos^2(3\theta)-\cos(3\theta)-2=0$

Next,  solve the quadratic using factorising:

$(2\cos(3\theta)+1)(3\cos(3\theta)-2)=0$

or $\cos(3\theta)=-\frac{1}{2}$ and $\cos(3\theta)=\frac{2}{3}.$

Note that if $0\textless\theta\textless180^\circ$, then $0\textless 3\theta\textless540^\circ$. Solutions on this interval can be seen in the graph below where we let $x=3\theta$:

If the x solutions are 48.19, 120, 240, 311.81, 408.19 and 480, then the solutions to the original equation are $\theta=16.06^\circ, 40^\circ, 80^\circ, 103.94^\circ, 136.06^\circ, 160^\circ$.

Note that when factorising a trigonometric quadratic, some factors do not produce solutions. For example, $\cos(x)=2$ does not have any solutions. This is because sin and cos only takes values between -1 and 1. Tan, however, is not restricted in this way but does have asymptotes where solutions are restricted – see the tan graph above (Example 1). In both cases above (Examples 3 and 4), both factors produced solutions.

Alternatively, click here to find Questions by Topic and scroll down to all past TRIGONOMETRY exam questions to practice some more.