Sin^2 x + cos^2 2x = 1

Решите задачу:

$\sin^2x+\cos^22x=1;\\ \sin^2x+(\cos2x)^2=1;\\ \sin^2x+(1-2\sin^2x)^2=1;\\ 4\sin^4x-4\sin^2x+1+\sin^2x=1;\\ 4\sin^4x-3\sin^2x=0;\\ \sin^2x(4\sin^2x-3)=0;\\ 1)\sin^2x=0;\\ \sin x=0;\\ x=\pi n, n\in Z;\\ 2)4\sin^2x-3=0;\\ \sin^2x=\frac34;\\ \sin x=\pm\frac{3}{2};\\$
$a) \sin x=+\frac{\sqrt3}{2};\\ x=(-1)^k\arcsin\frac{\sqrt3}{2}+\pi k\\ x=(-1)^k\frac\pi3+\pi k;\\ b)\sin x=\frac{\sqrt3}{2};\\ x=(-1)^m\arcsin(-\frac{\sqrt3}{2})+\pi m;\\ x=-(-1)^m\arcsin\frac{\sqrt3}{2}+\pi m;\\ x=-(-1)^m\frac\pi3+\pi m;\\ a)\bigcup b): x=\pm(-1)^k\cdot\frac{\pi}{3}+\pi k, k\in Z;\\ \\ \\ \\ \left[ {{x=\pi n,\ \ \ \ \ \ \ \ \ \ \ \ \ } \atop {x=-(-1)^k\cdot\frac{\pi}{3}+\pi k}} \right.\ \ \ n,k\in Z$