5 sin в квадрате x - 3 sinx*cos x-2cos в квадрате x=0

5sin^2x - 3sinxcosx - 2cosx = 0;

5sin^2x - 3sinxcox - cos^2x - sin^2x = 0 | : cos^2x;

5tg^2x - 3tgx - 1 - tg^2x = 0;

4tg^2x - 3tgx - 1 = 0;

Делаем замену tgx = t;

4t^2 - 3t - 1 = 0

D = b^2 - 4ac = 9 - 4 * 4 * (-1) = 25;

t1 = 3+5/8 = 1; t2 = 3-5/8 = - 1/4;

tgx = 1; tgx = - 1/4

x = p/4 + pk, k e z; x = arctg ( - 1/4) + pn, n e z;

ответ : x1 = p/4 + pk, k e z; x2 = arctg ( - 1/4) + pn, n e z.