Question

2sin2x-sin^2x=3cos^2x

Answer 1

2sin2x - sin²x = 3cos²x
-sin²x + 4sinxcosx - 3cos²x = 0       |·(-1)
sin²x - 4sinxcosx + 3cos²x = 0        |:cos²x
tg²x - 4tgx + 3 = 0
tg²x - 4tgx + 4 - 1 = 0
(tg²x - 2)² - 1² = 0
(tgx - 2 - 1)(tgx - 2 + 1) = 0
(tgx - 3)(tgx - 1) = 0
tgx = 3                         или             tgx = 3 
x = arctg3 + πn, n ∈ Z или             x = π/4 + πk, k ∈ Z 
Ответ: x = x = π/4 + πk, k ∈ Z; arctg3 + πn, n ∈ Z.