[tex]\displaystyle\bf\\Cos^{2} x-2Cosx Sinx=0\\\\Cosx\cdot\Big(Cosx-2Sinx\Big)=0\\\\\\\left[\begin{array}{ccc}Cosx=0\\Cosx-2Sinx=0\end{array}\right\\\\\\1)\\\\Cosx=0\\\\\\x=\frac{\pi }{2} +\pi n,n\in Z\\\\2)\\\\Cosx-2Sinx=0\\\\\\\frac{Cosx}{Cosx} -\frac{2Sinx}{Cosx} =0 \ \ ; \ \ Cosx\neq 0\\\\\\1-2tgx=0\\\\2tgx=1\\\\tgx=\frac{1}{2} \\\\x=arctg\frac{1}{2} +\pi n,n\in Z\\\\\\Otvet \ : \ \frac{\pi }{2} +\pi n,n\in Z \ \ ; \ \ arctg\frac{1}{2} +\pi n,n\in Z[/tex]

Объяснение:

cos²x-2cosxsinx=0

cosx(cosx-2sinx)=0

cosx=0 x=π/2+kπ,k∈z

cosx-2sinx=0

-2sinx= -cosx

-2tgx= -1

tgx=1/2

x=arctg(1/2)+kπ,k∈z

решение: х=π/2+kπ,k∈z

x=arctg(1/2)+kπ,k∈z