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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.