Найдите корни уравнения sin10xsin2x=sin8xsin4x

Принадлежащие промежутку

1/2 (cos (10x - 2x) - cos (10x + 2x)) = 1/2 (cos (8x - 4x) - cos (8x + 4x))

cos 8x - cos 12x = cos 4x - cos 12x

cos 8x - cos 4x = 0

- 2 sin ((8x - 4x) / 2) · sin ((8x + 4x) / 2) = 0

sin 2x · sin 6x = 0

sin 2x = 0 или sin 6x = 0

2x = πn 6x = πk

x = πn/2 x = πk/6

- π/6 ≤ πn/2 ≤ π/2 - π/6 ≤ πk/6 ≤ π/2

- π ≤ 3πn ≤ 3π - π ≤ πk ≤ 3π

- 1/3 ≤ n ≤ 1 - 1 ≤ k ≤ 3

n = 0 ⇒ x = 0 k = - 1 x = - π/6

n = 1 x = π/2 k = 0 x = 0

k = 1 x = π/6

k = 2 x = π/3

k = 3 x = π/2

Ответ: - π/6; 0; π/6; π/3; π/2