2sin ((3p/2) - x) cos ((p/2) - x) = sqrt (2) cosx

2sin (3π/2 - x) * cos (π/2 - x) = √2 * cosx;

-2cosx*sinx = √2 * cosx;

2 cosx*sinx + √2 * cosx = 0;

2 cosx * (sinx + √2 / 2) = 0;

a) cosx = 0 ⇒x = π/2 + π*n, n∈Z.

b) sinx + √2 / 2 = 0;

sinx = - √2 / 2;

x = (-1) ^ (n+1) * (π/4) + π*n, n∈Z ...

ответ : { π/2 + π*n; (-1) ^ (n+1) * (π/4) + π*n, n∈Z }