По формулам приведения:
sin (π/2 - x) = cos x
cos (3π/2 + x) = sin x
sin (π - x) = sin x
Получаем:
2cos x*sin x = sqrt(3)*sin x
2cos x*sin x - sqrt(3)*sin x = 0
sin x*(2cos x - sqrt(3)) = 0
1) sin x = 0
x1 = πn, n ∈ Z
2) 2cos x - sqrt(3) = 0
cos x = sqrt(3)/2
x2 = π/6 + 2πk, k ∈ Z
x3 = - π/6 + 2πk, k ∈ Z