Question

22sin^x-9sin2x=20 помогите, плиз.......

Answer 1

22 sin^2 - 18sin x* cos x = 20(sin^2 x + cos^2 x);
2 sin^2 x - 18sinx * cos x - 20 cos^2 x = 0;   /2cos^2 ≠ 0;
tg^2 x - 9 tg x - 10 = 0;
D = 81 - 4*(-10) = 121= 11^2;
 tg x = (9-11) /2 = - 1; ⇒ x = - pi/4 + pi*k; k-Z.
tg x = (9+11) / 2 = 10; ⇒x = frctg 10 + pi*k; k-Z.