2sin^2x-5sin 2x cos 2x+2cos^2x

2sin^2x-5sin 2x cos 2x+2cos^2x=0;2tq²2x - 5tq2x +2 =0 ; tq²2x - (1/2+2)tq2x +1 =0 ; [ tq2x =1/2 ; tq2x =2 . [ 2x=arctq(1/2) +πn ; [ 2x=arctq2 +πn ,  n∈Z . x₁=(1/2)*arctq(1/2) +(π/2)*n , n∈Z  ;  x₂=(1/2)*arctq2 +(π/2)*n,  n∈Z . ответ : (1/2)*arctq(1/2) +(π/2)*n , (1/2)*arctq2 +(π/2)*n,  n∈Z .