sin2x=sinx-2cosx+1
на отрезке 3pi/2;3pi

(2sinxcosx+2cosx)-(sinx+1)=02cosx(sinx+1)-(sinx+1)=0(sinx+1)(2cosx-1)=0sinx=-1x=-π/2+2πn,n∈z3π/2≤-π/2+2πn≤3π3≤-1+4n≤64≤4n≤71≤n≤7/4n=1⇒x=-π/2+2π=3π/2cosx=1/2x=-π/3+2πk,k∈z U x=π/3+2πm,m∈z3π/2≤-π/3+2πk≤3π9≤-2+12k≤1811≤12k≤2011/12≤k≤20/12k=1⇒x=-π/3+2π=5π/33π/2≤π/3+2πm≤3π9≤2+12m≤187≤12m≤167/12≤m≤16/12m=1πx=π/3+2π=7π/3