Finding fundamental period of $f(x)=\sec($sin(x)$)$

First observe that $\pi$ is a period of this function because $\sec (\sin (\pi +x))=\sec (-\sin \, x)=\sec (\sin \, x))$. Next you have to prove that no number in $(0,\pi)$ is a period. Let $0<p<\pi$. If possible suppose $\sec (\sin (p +x))=\sec (\sin \, x)$ for all $x$. Put $x=0$ to see that $\sec (\sin \, p )=1$. This means $\cos (\sin \, p )=1$. But $t=\sin \, p$ lies strictly between $0$ and $1$ so $\cos \, t$ cannot be $1$. This proves that $\pi$ is the fundamental period.

This is just a ruling out procedure assuming one of the answers is correct.

Consider

$f(x)=\cos (\sin x);$

$f(x+T)=\cos (\sin (x+T))=$

$\cos ( \sin x \cos T +\cos x \sin T);$

1) $T=π/2$:

$f(x+π/2)= \cos (\cos x);$

$f(x)= \cos (\sin x) \not =f(x+π/2)= \cos( \cos x).$

2) Check $T=π√$

3) $T= 2π$ is not the fundamental period (answer $2$)

4) Rule out .

Show that $T=π$ is the fundamental period.

Since $π$ is a period (possibly not fundamental) we conclude that any possible fundamental period needs to be of the form $π/k$, where $k >2$ , e.g. $π/3, π/5, π/7, π/9, ......$ (Rule out denominators with multiples of $2$(why?)).

Then

$\sin x=\sin x \cos π/k + \cos x \sin π/k$

$0 < \sin π/k =:b <1$, $0< \cos π/k =:a <1$.

$\sin x =a \sin x +b \cos x$;

This is impossible for $x \in \mathbb{R}$.

We conclude $T=π$ is the fundamental period.