Solve $\cos x-\sin(2x)=0$ Solve $\cos x-\sin(2x)=0$
I did:
$$\cos x=\color{blue}{\sin(\pi /2-x)}$$
therefore:
$$\color{blue}{\sin(\pi /2-x)}=\sin(2x)$$

Can I do that:??
now to solve only for $\pi/2-x=2x$
so $x=\pi/6+2\pi k$


Cos x -sin(2x)= 0   implies that cos x- 2 sin x cos x=0      cos x (1- 2sin x) = 0
It means   cos x=0  or sin x=1/2
   Which gives answer  x = 2n pi +- 90`    or x= n pi +(-1)^n pi/6    I hope it helps you   
