Помогите решить sin2x=корень из 3*cos(3pi\2-x)

Решениеsin2x = √3*cos(3pi\2-x)sin2x + √3sinx = 02sinxcosx + √3sinx = 0sinx(2cosx + √3) = 01) sinx = 0x₁ = πk, k∈Z2)  2cosx + √3 = 0cosx = - √3/2x = (+ -)arccos(-√3/2) + 2πn, n∈Zx = (+ -)(π - arccos√3/2) + 2πn, n∈Zx = (+ -)(π- π/6) + 2πn, n∈Zx₂ = (+ -)(5π/6) + 2πn, n∈Z