А) 2√3cos^2(3pi/2+x)-sin2x=0 б)отрезок [3pi/2;

(2√3)*[cos^2(3π/2 + x)] - sin2x = 0
(2√3)*[sin^2(x)] - sin2x = 0
(2√3)*[sin^2(x)] - 2sinxcosx = 0
sinx(√3sinx - cosx) = 0
1)  sinx = 0
x1 = πk, k∈Z
2)  √3sinx - cosx = 0   / cosx ≠ 0
√3tgx - 1 = 0
√3tgx = 1
tgx = 1/√3
x = arctg(1/√3) + πn, n∈Z
x2 = π/6 + πn, n∈Z