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 $2 \pi n$

Note that, if $\cos x >0$, $x$ is in the 1st or 4th quadrant; if $\cos x< 0$, $x$ is in the 2nd or 3rd quadrant. 
So, given $3\theta \in [-3\pi, 3\pi]$ and $\cos 3\theta = -\frac12< 0$, the angle $3\theta$ has two sets of values, with one set in the 2nd quadrant, i.e.
$$3\theta =\frac{2\pi}3 + 2\pi k = -\frac{4\pi}3 , \>\frac{2\pi}3, \>  \frac{8\pi}3$$
and the other set in the 3rd quadrant, i.e.
$$3\theta = -\frac{2\pi}3 + 2\pi k =-\frac{8\pi}3,\>  -\frac{2\pi}3, \>\frac{4\pi}3$$
Thus, all eligible angles are $ \pm\frac{2\pi}9, \> \pm\frac{4\pi}9, \> \pm\frac{8\pi}9$.
