Решите cosx+sin^2x=cos^2x

cosx+sin²x=cos²xcosx+1-cos²x-cos²x=02cos²x-cosx-1=0cosx=t2t²-t-1=0D=1+8=9=3²t=(1±3)/4t1=1;t2=-1/2cosx=1;x=2πncosx=-1/2x=±(π-π/3)+2πkx=±2π/3+2πk;k€Z