Решить уравнение
2cos^ 2x+sin2x-1=0

2cos^2 x - 1 = cos 2xcos 2x + sin 2x = 0√2*(cos 2x*1/√2 + sin 2x*1/√2) = 0√2*(sin(pi/4)*cos(2x) + cos(pi/4)*sin(2x)) = 0√2*sin (2x + pi/4) = 02x + pi/4 = pi*kx = -pi/8 + pi/2*k