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