Question

2sin^2x+3sinxcosx+7cos^2x=6

Answer 1

2sin^2x+3sinxcosx+7cos^2x=6*1
2sin^2x+3sinxcosx+7cos^2x=6*(Sin^2 x + Cos^2 x) 
2sin^2x+3sinxcosx+7cos^2x=6Sin^2x + 6Sin^2x
-4Sin^2x +3SunxCosx + Cos^2x= 0 | : Cos^2x
-4tg^2x +3tgx +1 = 0
tgx = z
-4z^2 +3z +1 = 0
D = 25
z₁ = -1/4;     z₂ = 1
a) tgx = -1/4                                   б) tgx = 1
x = -arctg1/4 + πk , k ∈Z                   x = π/4 + πn , n ∈ Z