Question

$6cos^2 x + 13sin 2x = -10$

Answer 1

6cos²x + 13*2sinxcosx = - 10(sin²x + cos²x)
6cos²x + 26sinxcosx = - 10sin²x - 10cos²x
6cos²x + 10cos²x + 26sinxcosx + 10sin²x = 0  
16cos²x + 26sinxcosx + 10sin²x = 0  /:2
8cos²x + 13sinxcosx + 5sin²x = 0  /:cos²x ≠ 0 
5tg²x + 13tgx + 8 = 0 

tgx = t 
5t² + 13t + 8 = 0 
D = 169 - 160 =  9
t1 = ( - 13  + 3)/10 = - 10/10 = - 1
t2 = ( - 13 - 3)/10 = - 16/10 = - 8/5

1) tgx = - 1
x = - pi/4 + pik,k ∈Z

2) tgx = - 8/5
x = - arctg(8/5) + pik, k ∈Z