Question

sin квадрат2x-cos квадрат2x=√3/2

Answer 1

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