Result comparison of a trigonometric equation The Problem: Solve the equation: $$ \cos x=\cos3x+ 2\sin2x
\\$$
The Result: $$ x=k\frac{\pi}{2},k\in \mathbb{Z}$$ 
My solution:$$ \cos x=4\cos^3x-3\cos{x}+4\sin{x}\cos{x}\\ 4\cos^3x-4\cos{x}+4\sin{x}\cos{x}=0\\ 4\cos{x}(\cos^2{x}+\sin{x}-1)=0\\
\cos{x}(1-\sin^2{x}+\sin{x}-1)=0\\
\cos{x}(\sin{x}-\sin^2{x})=0\\
\cos{x}\sin{x}(1-\sin{x})=0\\
x=\frac{\pi}{2}+l\pi \lor x=m\pi \lor x=\frac{\pi}{2}+2n\pi; \hspace{0.4cm}l,m,n\in \mathbb{Z}$$
Question: Is my solution equal to the result?

My resolution: 
$$\cos x= \cos \left(3x\right)+2\sin \left(2x\right)$$
$$-\cos \left(3x\right)-2\sin \left(2x\right)+\cos \left(x\right)=0$$
Using the following identity
$$-\cos \left(p\right)+\cos \left(q\right)=2\sin \left(\frac{p+q}{2}\right)\sin \left(\frac{p-q}{2}\right)$$
I obtaining,
$$-2\sin \left(2x\right)+2\sin \left(\frac{-x+3x}{2}\right)\sin \left(\frac{x+3x}{2}\right)=0$$
Hence
$$-2\sin \left(2x\right)+2\sin \left(2x\right)\sin \left(x\right)=0$$
Factorizing 
$$2\sin \left(2x\right)\left(\sin \left(x\right)-1\right)=0$$
Solving each part separately:
$$ \sin \left(2x\right)=0\quad \mathrm{or}\quad \sin \left(x\right)-1=0$$
I have your solutions:
$$\sin \left(2x\right)=0\quad :\quad x=k\pi,\:x=\frac{\pi}{2}+k\pi$$
$$\sin \left(x\right)-1=0\quad :\quad x=\frac{\pi}{2}+2k\pi,\, k\in\mathbb Z$$
