1+2cosx=sin2x+2sinx x€[-7/2π;-2π]
1 + 2Cosx - Sin2x - 2Sinx = 0
(1 - Sin2x) + ( 2Cosx -2Sinx) = 0
(Sin² x + Cos²x - 2SinxCosx) + 2(Cosx -Sinx) = 0
(Cosx -Sinx)² + 2(Cosx -Sinx) = 0
(Cosx - Sinx)(Cosx -Sinx +2) = 0
Cosx - Sinx = 0 |:Cosx или Cosx -Sinx +2 = 0
1 - tgx = 0 Cosx -Sinx = -2
tgx = 1 ∅
x = π/4 + πk , k ∈ Z