Writing answer for $2\cos x+\sin2x=0$ $$2\cos x + \sin 2x = 0\implies
\cos x = 0 \; \&\; \sin x = -1$$
So, the solution my book provides is
$x = π/2 + 2nπ\; \&\; x = 3π/2 + 2nπ$
Why is $3π/2$ (for $\cos$) not included in the general solution set?


$$2\cos x+\sin2x=0\\2 \cos x+ 2 \sin x \cos x=0 \\ \to 2 \cos x (1+\sin  x)=0\\ \to \\ sinx=-1 ,cosx=0$$
With respect to figure $$\begin{cases}cos=0 & x = \pm \frac{\pi}{2}+2k\pi\\sinx=-1 & x=-\frac{\pi}{2}+2k\pi\end{cases} $$ and $$x=\pm \frac{\pi}{2} +2k\pi$$ include $x=\frac{3\pi}{2} ,\frac{-3\pi}{2},\frac{\pi}{2},\frac{-\pi}{2},...$so we can rewrite it as  $\frac{3\pi}{2}+2k\pi \\or\\ \frac{-\pi}{2}+2k\pi$
