Sin2x=2Sinx-Cosx+1

Решить уравнение, найти корни на промежутке [-2 П; -П/2)

2sinxcosx-2sinx+cosx-1=0

2sinx (cosx-1) - (cosx-1) = 0

(cosx-1) (2sinx-1) = 0

cosx=1⇒x=2πk, k∈z

-2π≤2πk≤-π/2

-4≤4k≤-1

-1≤k≤-1/4

k=-1 x=-2π

sinx=1/2

x=π/6+2πk U x=5π/6+2πk, k∈z

-2π≤π/6+2πk≤-π/2

-12≤1+12k≤-3

-13/12≤k≤-4/12

k=-1 x=π/6-2π=-11π/6

-2π≤5π/6+2πk≤-π/2

-12≤5+12k≤-3

-17/12≤k≤-8/12

k=-1 x=5π/6-2π=-7π/6