Решите уравнение: 4 cos^2x-3 sinx=3

4cos^2x - 3sinx - 3 = 0

4 (1 - sin^2x) - 3sinx - 3 = 0

4 - 4sin^2x - 3sinx - 3 = 0

- 4sin^2x - 3sinx + 1 = 0 / / : (-1)

4sin^2x + 3sinx - 1 = 0

Пусть sinx = t, тогда

4t^2 + 3t - 1 = 0

Δ = 9 + 16 = 25

t1 = ( - 3 + 5) / 8 = 1/4

t2 = ( - 3 - 5) / 8 = - 1

sinx = 1/4

x = (-1) ^k arcsin (1/4) + pik, k ∈ Z

sinx = - 1

x = - π/2 + 2πn, n ∈ Z.