Помогите решить
Sin (pi/6 -x)*sinx=sin2x

sin (pi/6 - x)*sin x = sin 2x = 2sin x*cos x
sin x*(sin(pi/6 - x) - 2cos x) = 0
1) sin x = 0; x1 = pi*k
2) sin(pi/6 - x) - 2cos x = 0
sin(pi/6)*cos x - cos(pi/6)*sin x - 2cos x = 0
1/2*cos x - √3/2*sin x - 2cos x = 0
-√3/2*sin x - 3/2*cos x = 0
-√3*sin x = 3cos x
tg x = -3/√3 = -√3; x2 = -pi/3 + pi*k
Ответ: x1 = pi*k; x2 = -pi/3 + pi*k