2cos^2x + (2-√2)sinx+√2-2=0

cos^2x = 1 - sin^2x

2(1 - sin^2x) + (2-√2)sinx+√2-2=0

2-2sin^2x + (2-√2)sinx+√2-2=0

-2sin^2x+(2-√2)sinx+√2=0

2sin^2x-(2-√2)sinx-√2=0

D=(2-√2)^2 + 4*2*√2 = 4 - 4√2 + 2 + 8√2 = 6+4√2 = (√2+2)^2

sinx1=(2-√2 - √2+2) /4 = (4-2 √2)/4 = (2- √2)/2 = √2/2  ->x=(-1)^(n+1)   *pi/4+pi*n

sinx2 = (2-√2 +√2+2) /4 = 1  ->x=pi/2+2pi*k

как-то так, проверььте