# $-2(\sin x+2\cos 2x)=0$

I am finding the 2nd derivative critical values for graphing a trig function. So far I have it simplified to

$$-2(\sin x+2\cos 2x)=0$$

What values for x make this equal zero? And is there a reliable way to do this, besides staring at it until I figure it out...?

Use Double-Angle Formula, $$\cos2x=1-2\sin^2x$$ to form a Quadratic Eqaution in $\sin x$
Keep in mind that $\displaystyle-1\le\sin x\le1$ for real $x$