Sin2x + sin^2x + cos^2x = 1/2

$sin2x + sin^2x + cos^2x = \frac{1}{2} \\sin2x+(sin^2x+cos^2x)= \frac{1}{2} \\sin2x+1= \frac{1}{2} \\sin2x= \frac{1}{2} -1 \\sin2x=- \frac{1}{2} \\2x=- \frac{\pi}{6} +2\pi n \\x_1=- \frac{\pi}{12} +\pi n,\ n \in Z \\2x=- \frac{5\pi}{6} +2\pi n \\x_2=- \frac{5\pi}{12} +\pi n,\ n \in Z$
Ответ: $x_1=- \frac{\pi}{12} +\pi n,\ n \in Z ;\ x_2=- \frac{5\pi}{12} +\pi n,\ n \in Z$