Find the general solution to cos2x=-cosx

${\displaystyle \cos \left(2x\right)=2{\cos }^2\left(x\right)-1}$
${\displaystyle 2{\cos }^2\left(x\right)-1=\cos \left(x\right)}$
${\displaystyle 2{\cos }^2\left(x\right)-1-\cos \left(x\right)=0}$
${\displaystyle 2{\cos }^2\left(x\right)-\cos \left(x\right)-1=0}$
${\displaystyle (2\cos \left(x\right)+1)(\cos \left(x\right)-1)=0}$
${\displaystyle 2\cos \left(x\right)+1=0}$
${\displaystyle 2\cos \left(x\right)=-1}$
${\displaystyle \cos \left(x\right)=-\frac{1}{2}}$
${\displaystyle x={\cos }^{-1}(-\frac{1}{2})}$
$x=\pi /3[2\pi -\pi /3=5\pi /3]$
$x=\pi /3and5\pi /3$
cos is positive in the first and fourth quadrant.
${\displaystyle \cos \left(x\right)-1=0}$
${\displaystyle \cos \left(x\right)=1}$
${\displaystyle x={\cos }^{-1}\left(1\right)}$
${\displaystyle x=0,2\pi }$