# Find the general solution to cos2x=-cosx

Find the general solution to $\mathrm{cos}2x=-\mathrm{cos}x$
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$\mathrm{cos}\left(2x\right)=2{\mathrm{cos}}^{2}\left(x\right)-1$
$2{\mathrm{cos}}^{2}\left(x\right)-1=\mathrm{cos}\left(x\right)$
$2{\mathrm{cos}}^{2}\left(x\right)-1-\mathrm{cos}\left(x\right)=0$
$2{\mathrm{cos}}^{2}\left(x\right)-\mathrm{cos}\left(x\right)-1=0$
$\left(2\mathrm{cos}\left(x\right)+1\right)\left(\mathrm{cos}\left(x\right)-1\right)=0$
$2\mathrm{cos}\left(x\right)+1=0$
$2\mathrm{cos}\left(x\right)=-1$
$\mathrm{cos}\left(x\right)=-\frac{1}{2}$
$x={\mathrm{cos}}^{-1}\left(-\frac{1}{2}\right)$
$x=\pi /3\left[2\pi -\pi /3=5\pi /3\right]$

cos is positive in the first and fourth quadrant.
$\mathrm{cos}\left(x\right)-1=0$
$\mathrm{cos}\left(x\right)=1$
$x={\mathrm{cos}}^{-1}\left(1\right)$
$x=0,2\pi$