sin квадрат2x-cos квадрат2x=√3/2

Решениеsin²2x - cos²2x = √3/21 - 2cos²2x = √3/2cos4x = - √3/24x = (+ -)arccos(-√3/2) + 2πk, k ∈ Z4x = (+ -) (π - arccos(√3/2)) + 2πk, k ∈ Z4x = (+ -) (π - π/6) + 2πk, k ∈ Z4x = (+ -) (5π/6) + 2πk, k ∈ Zx = (+ -) (5π/24) + πk/2, k ∈ Z