а)
2sin²x-3√2cos(3π/2+x)+2=0
2sin²x-3√2sinx+2=0
D=18-16=2
(sinx)₁=(3√2-√2)/4=√2/2 ⇒ x=π/4+2πk; x=3π/4+2πk; k∈Z
(sinx)₂=(3√2+√2)/4=√2 ∉ [-1; 1] ⇒ ∅
б)
3π/4+2π=11π/4 ∈ [5π/2; 4π]
x=π/4+2πk ∉ [5π/2; 4π]
ответ: а) x=π/4+2πk; x=3π/4+2πk; k∈Z, б) 11π/4
