Корень из 2*sin (3 п/2-x) * cos (3 п/2+x) = cosx

1. По формулам приведения получим:

√2sin (3π/2 - x) * cos (3π/2 + x) = cosx; √2sin (π + π/2 - x) * cos (π + π/2 + x) = cosx; - √2sin (π/2 - x) * (-1) * cos (π/2 + x) = cosx; √2sin (π/2 - x) * cos (π/2 + x) = cosx; √2cosx * (-sinx) = cosx; - √2cosx * sinx = cosx; √2cosx * sinx + cosx = 0; cosx (√2sinx + 1) = 0.

2. Приравняем множители к нулю:

[cosx = 0;

[√2sinx + 1 = 0; [cosx = 0;

[√2sinx = - 1; [cosx = 0;

[sinx = - √2/2; [x = π/2 + πk, k ∈ Z;

[x = - π/4 + πk, k ∈ Z.

Ответ: π/2 + πk; - π/4 + πk, k ∈ Z.