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

Question

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

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 $2 \pi n$

Answer 1

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$.

Answer 2

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.

Answer 3

$\cos^{-1}(x)$’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 $\frac{2\pi}{9}-\frac{6\pi}{9}$.