Sin квадрат2x-cos квадрат2x=√3/2
Решение sin²2x - cos²2x = √3/2 1 - 2cos²2x = √3/2 cos4x = - √3/2 4x = (+ -)arccos(-√3/2) + 2πk, k ∈ Z 4x = (+ -) (π - arccos(√3/2)) + 2πk, k ∈ Z 4x = (+ -) (π - π/6) + 2πk, k ∈ Z 4x = (+ -) (5π/6) + 2πk, k ∈ Z x = (+ -) (5π/24) + πk/2, k ∈ Z