Question

Cosx-Cos2x=0
Solve. I'm not sure if I'm getting the right answers

Answer 1

`cos(2x)= 2cos^2(x)-1\\ \\ cos(x)-cos(2x)=0 \\ cos(x)-(2cos^2(x)-1)=0 \\ -2cos^2(x)+cos(x)+1=0 \\ 2cos^2(x)-cos(x)-1=0 \\ (2cosx+1)(cosx-1)=0 \\ \\ 2cos(x)+1=0 \\ cos(x)= - (1)/(2) \\ x= \pm arccos(- (1)/(2) )+2 \pi k~~~k\in Z \\ arccos(- (1)/(2) )= (2 \pi )/(3) \\ x=\pm (2 \pi )/(3)+2 \pi k~~~k\in Z \\ \\`



`cos(x)-1=0 \\ cos(x)=1 \\ x= 2\pi k~~~k \in Z`