Решите cosx+sin^2x=cos^2x
Cosx+sin²x=cos²x cosx+1-cos²x-cos²x=0 2cos²x-cosx-1=0 cosx=t 2t²-t-1=0 D=1+8=9=3² t=(1±3)/4 t1=1;t2=-1/2 cosx=1;x=2πn cosx=-1/2 x=±(π-π/3)+2πk x=±2π/3+2πk;k€Z