Sin6x+sin2x+2sin^2x=1

Решите задачу:

$\sin6x+\sin 2x+2\sin^2x=1 \\ 2\sin \frac{6x+2x}{2} \cdot \cos \frac{6x-2x}{2} +2\cdot \frac{1+\cos 2x}{2} =1 \\ 2\sin 4x\cos 2x+\cos 2x=0 \\ \cos 2x(2\sin 4x+1)=0 \\ \cos 2x=0 \\ 2x= \frac{\pi}{2} +\pi n ,n \in Z \\ x= \frac{\pi}{4} + \frac{\pi n}{2} , n \in Z \\ \sin 4x=- \frac{1}{2} \\ 4x=(-1)^{k+1}\cdot \frac{\pi}{6} +\pi k,k \in Z \\ x=(-1)^{k+1}\cdot \frac{\pi}{24}+ \frac{\pi k}{4} , k \in Z$