Let $T_n(x)$ be the $nth$ Chebyshev Polynomial. Prove that $|T_n(x)| \leq 1$ for $x \in [0,1]$ and determine when there is equality.
So far I was thinking that $t$ would have to be of the form $\frac{p}{q}\pi$ where $p,q\in \mathbb{Q}$ and $p<q$. Then the number of solutions would be $\lfloor{n/q}\rfloor$.
For $-1 \le x \le 1$ the trigonometric definition of the Chebyshev polynomials is $$ T_n(x) = \cos(n \arccos(x)) \, , $$ which immediately implies $|T_n(x)| \le 1$ in that range.
If we exclude the (trivial) case $n=0$ then equality holds exactly if $n \arccos(x)$ is an integer multiple of $\pi$, i.e. if $$ x = \cos(\frac {k \pi}n) $$ for some $k \in \Bbb Z$. It remains to determine which values of $k$ lead to solutions $x \in [0, 1]$.