2sin²x+3sin2x+2=0 Решить через пусть

2sin^2 (x) + 3sin (2x) + 2 = 0;

2sin^2 (x) + 6sin (x) cos (x) + 2sin^2 (x) + 2cos^2 (x) = 0;

4sin^2 (x) + 6sin (x) cos (x) + 2cos^2 (x) = 0;

Разделим на 2cos^2 (x) ≠ 0; x ≠ п/2 + пn, n∈Z.

2tg^2 (x) + 3tg (x) + 1 = 0;

tg (x) = t;

2t^2 + 3t + 1 = 0;

t1 = (-3 + 1) / 4 = - 1/2;

t2 = (-3 - 1) / 4 = - 1;

1) tg (x) = - 1/2 = => x = - arctg (1/2) + пn, n∈Z.

2) tg (x) = - 1 = => x = - п/4 + пk, k∈Z.

Ответ: x = - arctg (1/2) + пn, n∈Z или - п/4 + пk, k∈Z.