Question

4 sin²x + 4 sinx+1=0​

Answer 1

Объяснение:

4 sin²x + 4 sinx+1=0

t = sin x

4t² + 4t + 1 = 0

D = 4^2 - 4 * 4 * 1 = 16 - 16 = 0

t = -4 / (2*4) = -1/2

sin x = -1/2

x = (-1)^(n+1) * (π/6) + πn , n ∈ Z