2· (1 - sin²x) + 1 = - 2√2sinx
-2sin²x + 3 + 2√2sinx = 0
2sin²x - 2√2sinx - 3= 0
sinx = t
2t² - 2√2t - 3 = 0
D = 8 + 24 = 32
t = (2√2 + 4√2)/ 4 = 3√2/2 или t = (2√2 - 4√2)/ 4 = - √2/2
sinx= 3√2/2 sinx = - √2/2
нет корней, т.к. 3√2/2 >1 x = (-1)^(n+1) π/4 +πn, n∈Z