Решите уравнения[tex]2\sin^2x-3\sin x+1=0;\\\tan^2x+2 ctg(\pi/2+x)+1=0.[/tex]

2sin²x-3sinx+1=0sinx=a2a²-3a+1=0D=9-8=1a1=(3-1)/4=1/2⇒sinx=1/2⇒x=(-1)^n*π/6+πn,n∈za2=(3+1)/4=1⇒sinx=1⇒x=π/2+2πk,k∈ztg²x-2tgx+1=0(tgx-1)=0tgx-1=0tgx=1x=π/4+πn,n∈z