Question

2sin квадратx-sinx-1=0

Answer 1

$2sin^2x-sinx-1=0 \\sinx=y,\ y \in [-1;1] \\2y^2-y-1=0 \\D=1+8=9=3^2 \\y_1=\frac{1+3}{4}=1 \\y_2=\frac{1-3}{4}=-\frac{1}{2} \\sinx=1 \\x_1=\frac{\pi}{2}+2\pi n,\ n \in Z \\sinx=- \frac{1}{2} \\x_2=- \frac{\pi}{6} +2\pi n,\ n \in Z \\x_3=- \frac{5\pi}{6} +2\pi n,\ n \in Z$