Finding all angles that satisfy $8 \cos ^{3} \theta-6 \cos \theta+1=0 \quad \text { for } \theta \in[-\pi, \pi]$ 
$\text { Hence, solve the equation } 8 \cos ^{3} \theta-6 \cos \theta+1=0 \quad \text { for } \theta \in[-\pi, \pi]$

The previous part was to prove that $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta \quad \text { by replacing } 3 \theta \text { by }(2 \theta+\theta)$. 
So, I used this to simplify the equation to 
$2 \cos 3 \theta +1 = 0$
$\implies \cos 3 \theta =\frac{-1}{2}$
Since $\cos^{-1} \frac{-1}{2} = \frac{2\pi}{3} + 2 \pi n$,$\implies \theta =\frac{2 \pi}{9},\frac{8 \pi}{9},\frac{-8 \pi}{9}$ or $\frac{-2 \pi}{9}$. However, the graph seems to be showing another root which is $\frac{4 \pi}{9}$. Why did I miss this root? How should I find more angles that satisfy the equation in a given range. In general, how does one find all angles that satisfy an equation even after adding $2 \pi n$

$\cos^{-1}(x)$’s range is $[0 , \pi]$ so if we calculate $\cos^{-1}(-\frac{1}{2})$ in our calculator, we are going to get $\frac{2\pi}{3}$
However, the solution to $\cos(x)=-\frac{1}{2}$ are $\pm\frac{2\pi}{3}+2n\pi$ with integer $n$. Since $x=3\theta$, $\theta=\pm\frac{2\pi}{9}+n\frac{6\pi}{9}$.
You missed $\theta=-\frac{2\pi}{9}+\frac{6\pi}{9}$ and $\frac{2\pi}{9}-\frac{6\pi}{9}$.
