<s>[INST] 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 $2 \pi n$
 [/INST] By your work $$\theta=\pm40^{\circ}+120^{\circ}k,$$ where $k\in\mathbb Z,$ which gives $$\{\pm40^{\circ},\pm80^{\circ},\pm160^{\circ}\}.$$
We see these roots in your picture. 
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