# Solve for $x$ given $\sin x$ and $2\cos(2x)$

I am trying to find absolute maximum and absolute minimum values for$\ f(x)=\cos(x)+\sin(2x)$ on the interval $\ [0,\fracπ2]$.

I started off by finding the derivative:

$\ f'(x)=-\sin x+2\cos(2x)$

Then I tried to find the critical numbers. But that's where I ran amuck. I came up with:

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

How do I go about solving for x?

use that $$\cos(2x)=1-2\sin^2 (x)$$ plugging this in your equation we get $$-\sin(x)+2(1-2\sin^2 (x))=0$$ this is a quadratic equation in $\sin(x)$