4sin ^2 x + 4 cos x - 1=0

Решение

4sin∧2x + 4cosx - 1 = 0

4 * (1 - cos∧2x) + 4 cosx - 1 = 0

4 - 4cos∧2x + 4cosx - 1 = 0

4cos∧2x - 4cosx - 3 = 0

D = 16 + 4*4*3 = 64

1) cosx = (4 - 8) / 8 = - 1/2

cosx = - 1/2

x = ( + - ) arccos (-1/2) + 2πn, n∈Z

x = ( + - ) (π - π/3) + 2πn, n∈Z

x = ( + - ) (2π/3) + 2πn, n∈Z

2) cosx = (4 + 8) / 8 = 3/2 не удовлетворяет области определения функции y = cosx (-1 ≤ cosx ≤ 1)