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 $x=\pi/6+2\pi k$


The first step is ok, but for the second we have that
$$\sin A=\sin B\iff A+B=(2k+1)\pi\text{ or }A-B=2k\pi$$
Then, from the equality $\sin(\pi/2-x)=\sin 2x$ we get two sets of solutions:


*

*$\pi/2-x+2x=(2k+1)\pi$

*$2x-(\pi/2-x)=2k\pi$

