Solve for x: 2cos^2 x+3cos x + 1=0

b) Factor the left side of the equation.

Let ${\displaystyle u=\cos \left(x\right)}$. Substitute ${\displaystyle u}$ for all occurrences of ${\displaystyle \cos \left(x\right)}$.

${\displaystyle 2u^2+3u+1=0}$

Factor by grouping.

${\displaystyle (2u+1)(u+1)=0}$

Replace all occurrences of ${\displaystyle u}$ with ${\displaystyle \cos \left(x\right)}$.

${\displaystyle (2\cos \left(x\right)+1)(\cos \left(x\right)+1)=0}$

If any individual factor on the left side of the equation is equal to 0${\displaystyle 0}$, the entire expression will be equal to ${\displaystyle 0}$.

${\displaystyle 2\cos \left(x\right)+1=0}$

${\displaystyle \cos \left(x\right)+1=0}$

Set ${\displaystyle 2\cos \left(x\right)+1}$ equal to ${\displaystyle 0}$ and solve for ${\displaystyle x}$.

Set ${\displaystyle 2\cos \left(x\right)+1}$ equal to ${\displaystyle 0}$.

${\displaystyle 2\cos \left(x\right)+1=0}$

Solve ${\displaystyle 2\cos \left(x\right)+1=0}$ for ${\displaystyle x}$.

${\displaystyle x=\frac{2\pi }{3}+2\pi n,\frac{4\pi }{3}+2\pi n}$, for any integer ${\displaystyle n}$

Set ${\displaystyle \cos \left(x\right)+1}$ equal to ${\displaystyle 0}$ and solve for ${\displaystyle x}$.

Set ${\displaystyle \cos \left(x\right)+1}$ equal to ${\displaystyle 0}$.

${\displaystyle \cos \left(x\right)+1=0}$

Solve ${\displaystyle \cos \left(x\right)+1=0}$ for ${\displaystyle x}$.

${\displaystyle x=\pi +2\pi n}$, for any integer ${\displaystyle n}$

The final solution is all the values that make ${\displaystyle (2\cos \left(x\right)+1)(\cos \left(x\right)+1)=0}$ true.

${\displaystyle x=\frac{2\pi }{3}+2\pi n,\frac{4\pi }{3}+2\pi n,\pi +2\pi n}$, for any integer n