Решить sin2x=корень из 3*cos(3pi\2-x)

Решение
sin2x = √3*cos(3pi\2-x)
sin2x + √3sinx = 0
2sinxcosx + √3sinx = 0
sinx(2cosx + √3) = 0
1) sinx = 0
x₁ = πk, k∈Z
2)  2cosx + √3 = 0
cosx = - √3/2
x = (+ -)arccos(-√3/2) + 2πn, n∈Z
x = (+ -)(π - arccos√3/2) + 2πn, n∈Z
x = (+ -)(π- π/6) + 2πn, n∈Z
x₂ = (+ -)(5π/6) + 2πn, n∈Z