Find a general formula for the points $x$ at which $T_n=\pm 1$. How many such points are there on $[-1,1]$? Hint: begin with a special case such as $n=3$.
We know that $|T_n(x)|\leq 1$ on $[-1,1]$ and $T_n(1)=1$.
My work: $T_n(x)=\cos(n\cos^{-1}(x))=\pm1 \text{ on } [-1,1].$ We can solve this by solving for $\cos(n\theta)=\pm1, \theta=\cos^{-1}(x)$
$$\begin{aligned}[c]
\cos(\theta)=1\\
\theta =\pm2k\pi\\
x=\cos\left(\frac{2k\pi}{n}\right)
\end{aligned}
\qquad\qquad
\begin{aligned}[c]
\cos(\theta)=-1\\
\theta=\pm\pi+2k\pi\\
x=\cos\left(\frac{\pi}{n}+\frac{2k\pi}{n}\right)
\end{aligned}$$
However, the solution I found from a different source is $$x=\cos\left(\frac{k\pi}{n}\right)$$ Does my answer just have to be simplified or is my method to solve this wrong? Furthermore, is there not an infinite number of $x's$ that will solve $T_n(x)$ since there is infinite number of $n's$ to choose from?