Question

How do you solve `sin(2x)=cos(x)`?

Answer 1

`x=-\pi/2,\pi/6,\pi/2, or {5\pi}/6` plus multiples of `2\pi`.

Explanation:

I cannot tell whether you have learned the double-angle formulas yet so I will develop a solution without them.

First observe that `\cos u=\cos v` when `u+v=2n\pi` or `u-v=2n\pi` for any integer `n`.

We then have `\sin(2x)=\cos(\pi/2-2x)`. So:

`\cos(\pi/2-2x)=\cos(x)`.

Then we have the two possibilities defined above.

Possibility 1:

`u+v=(\pi/2-2x)+x=2n\pi`
`\pi/2-x=2n\pi`
`x=\pi/2-2n\pi`

So `x=\pi/2` plus multiples of `2\pi`

Possibility 2:

`u-v=(\pi/2-2x)-x=2n\pi`
`\pi/2-3x=2n\pi`
`x=\pi/6-{2n}/{3\pi}`

So `x=-\pi/2, \pi/6` or `{5\pi}/6` plus multiples of `2\pi`.

Put the two possibilities together to get the answer.