Question

Sin2x=sinx-2cosx+1 на отрезке 3pi/2;3pi

Answer 1

(2sinxcosx+2cosx)-(sinx+1)=0
2cosx(sinx+1)-(sinx+1)=0
(sinx+1)(2cosx-1)=0
sinx=-1
x=-π/2+2πn,n∈z
3π/2≤-π/2+2πn≤3π
3≤-1+4n≤6
4≤4n≤7
1≤n≤7/4
n=1⇒x=-π/2+2π=3π/2
cosx=1/2
x=-π/3+2πk,k∈z U x=π/3+2πm,m∈z
3π/2≤-π/3+2πk≤3π
9≤-2+12k≤18
11≤12k≤20
11/12≤k≤20/12
k=1⇒x=-π/3+2π=5π/3
3π/2≤π/3+2πm≤3π
9≤2+12m≤18
7≤12m≤16
7/12≤m≤16/12
m=1πx=π/3+2π=7π/3