Решение
(sin2x * 2cos²x) / √sinx = 0
sin2x*2cos²x = 0
√sinx ≠ 0
sin2x = 0
2cos²x = 0
sinx ≠ 0, x ≠ πm, m∈Z
1) sin2x = 0
2x = πk, k∈Z
x = πk/2, k∈Z
2) cos²x = 0
(1 + cos2x)/2 = 0
cos2x = - 1
2x = π + 2πn, n∈Z
x = π/2 + πn, n∈Z
Ответ: x₁ = πk/2, k∈Z ; x = π/2 + πn, n∈Z