Решите уравнение 2-3*sin(x/2)*ctg(x/2)=sin^2(x/2)-sin^2(x/4) 99

2 - 3·sin(x/2)·ctg(x/2) = sin²(x/2) - sin²(x/4)

2 - 3·cos(x/2) = sin²(x/2) - 0,5(1 - cos(x/2))

2 - 3·cos(x/2) = 1 - cos²(x/2) - 0,5 + 0,5cos(x/2)

- 3·cos(x/2) =  - cos²(x/2) - 1,5 + 0,5cos(x/2)

cos²(x/2)  - 3,5cos(x/2) + 1,5 = 0|·2

2cos²(x/2)  - 7cos(x/2) + 3 = 0

Замена: cos(x/2) = t/2

t² - 7t + 6 = 0;

t₁ = 6; t₂ = 1.

Обратная замена:

cos(x/2) = 6/2 = 3 - не имеет решений

или

cos(x/2) = 1/2

x/2 = ±arccos(1/2) + 2πn, n∈Z;

x/2 = ±π/3 + 2πn, n∈Z;

x = ±2π/3 + 4πn, n∈Z.

ответ: ±2π/3 + 4πn, n∈Z.