16cos^4x-24cos^2x+9=0

Cos² x = t (*)

16 t ² - 24 t + 9 = 0

(4 t - 3) ² = 0

4t - 3 = 0

4t = 3

t = 3/4

Подставим в (*)

Сos² x = 3/4

a) Cos x = √3/2 или б) Cos x = - √3/2

x = + - arcCos √3/2 + 2πk, k ∈Z x = = - arcCos (-√3/2) + 2πn, n∈Z

x = + - π/6 + 2πk, k∈Z x = + - 5π/6 + 2πn, n ∈Z