Question

2cos^2x-3√3cosx+3=0

помогите пожалуйста...​

Answer 1

вот:

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