2sin ^ 2 (3п / 2-x) = cosx
[3п / 2; 0]

2сcos²x-cosx=0cosx*(2cosx-1)=0cosx=0⇒x=π/2+πn,n∈z0≤π/2+πn≤3π/20≤1+2n≤3-1≤2n≤2-1/2≤n≤1n=0⇒x=π/2n=1⇒x=π/2+π=3π/22cosx-1=0cosx=1/2x=-π/3+2πk,k∈z U x=π/3+2πt,t∈z0≤-π/3+2πk≤3π/20≤-2+12k≤92≤12k≤111/6≤k≤11/12нет решения0≤π/3+2πt≤3π/20≤2+12t≤9-2≤12t≤7-1/6≤t≤7/12t=0⇒x=π/3