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


Also, $$\cos x =\sin 2x \Rightarrow \cos x(1-2\sin x)=0 $$ $$ \cos x=0 \mid \sin x=\frac{1}{2}$$ where $\mid$ stands for "or". This can be easily solved.   
Also, you can see that $\frac{5\pi}{6}$ is clearly a solution that is not included by your approach. Hope it helps.

EDIT: Expanding on @ajotatxe's one-liner on why $\sin x= \sin y \nRightarrow x=y$:  
If $$\sin x-\sin y=0 \Rightarrow 2\cos \frac{x+y}{2}\sin \frac{x-y}{2} =0$$ Then solving, we have $$\frac{x+y}{2} = (2n+1)\frac{\pi}{2} \space\space\space\text{or}\space\space \frac{x-y}{2} = n\pi \space\space\space\text{where}\space\space n\in \mathbb Z$$ $$\Rightarrow x=(2n+1)\pi-y \space\space\space \text{or} \space\space x= 2n\pi + y \space\space\space\text{where}\space\space n\in \mathbb Z$$
