Вопрос по алгебре:

а)2sin^2x-3√2cos(3pi/2+x)+2=0 б)[5pi/2:4pi]

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