Sin2x=sinx-2cosx+1

на отрезке 3pi/2; 3pi

(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