Question

Sin2x=2Sinx-Cosx+1
Решить уравнение, найти корни на промежутке [-2П;-П/2)

Answer 1

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