Question

10 sin^2x+11 sin2x+6cos2x=-6

Answer 1

10 sin^2x + 11 sin2x + 6cos2x = - 6
10 sin^2x + 22 sinx cosx + 6(cos^2x - sin^2x) = - 6* 1
10 sin^2x + 22 sinx cosx + 12cos^2x =   0  /:cos^2x ≠ 0
10 tg^2x + 22 tgx + 12 = 0
tgx = t
10 t + 22t + 12 =0
D = 484 - 480 = 4
t = ( - 22 + 2)/ 20 = - 20/20 = - 1
t = ( - 22 - 2)/ 20 = - 24/20 = - 6/5

1) tgx = - 1 ==> x = - pi/4 + pik, k ∈Z
2) tgx = - 6/5 ===> x = - arctg(6/5) + pik, k ∈Z

ОТВЕТ:
x = - pi/4 + pik, k ∈Z
x = - arctg(6/5) + pik, k ∈Z