Result comparison of a trigonometric equation

Question

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?

Answer 1

Yes the condition

$$x=\frac{\pi}{2}+l\pi \lor x=m\pi \lor x=\frac{\pi}{2}+2n\pi; \hspace{0.4cm}l,m,n\in Z$$

is equivalent to

$$x=k\frac{\pi}{2},k\in Z$$

Answer 2

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: $$\sin \left(2x\right)=0\quad :\quad x=k\pi,\:x=\frac{\pi}{2}+k\pi$$ $$\sin \left(x\right)-1=0\quad :\quad x=\frac{\pi}{2}+2k\pi,\, k\in\mathbb Z$$