2 sinx * cosx - 7cos ^2x = 2 sin^2x

Question

2 sinx * cosx - 7cos ^2x = 2 sin^2x

Answer 1

2sin²x-2sinxcosx+7cos²x=0/cos²x≠0
2tg²x-2tgx+7=0
tgx=a
2a²-2a+7=0
D=4-56=-52<0⇒а-любое<br>tgx-любое
x∈(-π/2+πn;π/2+πn)