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

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