Pick all the roots of the equation 2sin2xctgx-sin ^ 2x = 3cosx + 1 belonging to the segment [0; pi].

We transform the equation by applying the double angle formula:
4 * sin (x) * cos (x) * cos (x) / sin (x) – sin ^ 2 (x) = 3 * cos (x) + 1;
4 cos ^ 2 (x) – (1 – cos ^ 2 (x)) – 3 * cos (x) – 1 = 0;
5 * cos ^ 2 (x) – 3 * cos (x) – 2 = 0;
We got a quadratic equation, find the roots:
cos (x) = [3 ± sqrt (9 + 4 * 5 * 2)] / 10 = (3 ± 7) / 10;
cos (x) = 1;
x = 2 * π * n;
cos (x) = -2/5;
x = ± arccos (-0.4) + 2 * π * n;
Find the roots belonging to the interval [0; π]
x1 = 0;
x2 = π – arccos (0.4).