Solving polynomials using De Moivre's theorem Given $\cos 4\theta=8\cos^4\theta−8\cos^2\theta+1$, solve $16x^4-16x^2+1=0$.
The textbook's answers are $x=\pm\cos\dfrac{\pi}{12},\pm\cos\dfrac{5\pi}{12}$. I managed to get two of the four answers and i can not figure out what i did wrong.
My Attempt
Let $x=\cos\theta$
$16\cos^4\theta-16\cos^2\theta+1=0$
$2(8\cos^4\theta-8\cos^2\theta+1)=1$
$2\cos4\theta=1$
$\cos4\theta=\dfrac{1}{2}$
$\cos4\theta=\cos\dfrac{\pi}{3}$
$\therefore4\theta=\dfrac{\pi}{3}+k\pi$, for any integer k
I chose $0\le k \le 3$ as the other solutions would repeat its values for x.
$\theta=\dfrac{\pi}{12},\dfrac{\pi}{3},\dfrac{7\pi}{12},\dfrac{5\pi}{6}$
$\therefore x=\cos\dfrac{\pi}{12},\cos\dfrac{\pi}{3},\cos\dfrac{7\pi}{12},\cos\dfrac{5\pi}{6}$
$x=\cos\dfrac{\pi}{12},\cos\dfrac{\pi}{3},-\cos\dfrac{5\pi}{12},\cos\dfrac{5\pi}{6}$

You wrote:
$$\cos4θ=\cfrac{1}{2}$$
$$\cos4θ=cos\cfrac{\pi}{3}$$
$$\therefore4θ=\cfrac{\pi}{3}+k\pi$$
This contains some errors
The $\cos$ and the $\sin$ function take each value to times in the interval $[0,2\pi]$ except $1$ and $-1$.
So the following holds:
$$\cos \phi=\cos (2\pi-\phi)$$
The period of these trigonometric functions is $2\pi$, and not $\pi$
Therefore you final line is
$$\therefore4θ=\pm\cfrac{\pi}{3}+2k\pi$$
After removing the duplicates four solutions remain.
