2 cos в кв x/3 + 3cos x/3-2=0

1) 2*cos^2 (x/3) + 3*cos (x/3) - 2 = 0

Замена: cos (x/3) = t, - 1 ≤ t ≤ 1

2t^2 + 3t - 2 = 0

D = 9 + 4*2*2 = 9 + 16 = 25

t1 = (-3 + 5) / 4 = 2/4 = 1/2

t2 = (-3 - 5) / 4 = - 8/4 = - 2 < - 1 - посторонний корень

cos (x/3) = 1/2

(x/3) = + - π/3 + 2πk

x = + - π + 6πk

2) Замена: cos (x/2) = t, - 1≤ t ≤ 1

2t^2 + √3*t = 0, t * (2t + √3) = 0

t1 = 0

t2 = - √3/2

cos (x/2) = 0, (x/2) = 2πk, x = 4πk

cos (x/2) = - √3/2, x/2 = + - 5π/6 + 2πk, x = + - 5π/3 + 4πk