Cos2x-cos^x+cos (0,5п+2x)=0

Question

Cos2x-cos^x+cos (0,5п+2x)=0

Answer 1

Сos²x-sin²x-cos²x-sin2x=0
sin²x+2sinxcosx=0
sinx*(sinx+2cosx)=0
sinx=0⇒x=πk,k∈z
sinx+2cosx=0 /cosx
tgx+2=0⇒tgx=-2⇒x=-arctg2+πk,k∈z