1 - 2sin²4x - 5sin4x - 3 = 0
-2sin²4x - 5sin4x - 2 = 0
2sin²4x + 5sin4x + 2 = 0
Пусть t = sin4x, t ∈ [-1; 1].
2t² + 5t + 2 = 0
D = 25 - 2•2•4 = 9 = 3²
t1 = (-5 + 3)/4 = -2/4 = -1/2
t2 = (-5 - 3)/4 = -8/4 = -2 - постронний корень
Обратная замена:
sinx = -1/2
x = (-1)ⁿ+¹π/6 + πn, n ∈ Z