# Trigonometric Equations

Example 1 – Solve the following for $0\leq x\leq 2\pi$:

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

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 must 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), or by using the inverse sine button on your calculator as well as the graph, we can see that the solutions on the interval $0\leq x\leq 2\pi$ are $0,\pi,2\pi$ and $\frac{\pi}{2}$.

Example 2 – Solve the following for $-\pi\leq x\leq \pi$:

$\cos(2x)=1$

It is very important to note here that if x can be between $-\pi$ and $\pi$ then 2x can be between $-2\pi$ and $2\pi$. When looking at the graph of cos to find values where cos takes the value 1, it is important to check all values between $-2\pi$ and $2\pi$.

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

$x=-\pi$, $x=0$ and $x=\pi$.

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.