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 }