I want to find the periodicity of the following function
$f(x) = \sqrt{\sin(x)}+\sqrt{\cos(x)}$
my attempt: I've posed $f(x)=f(x+T)$.So $$ \begin{align} f(x+T) &= \sqrt{\sin(x+T)}+ \sqrt{\cos(x+T)}\\ &= \sqrt{\sin(x)\cos(T)+\cos(t)\sin(T)}+\sqrt{\cos(x)\cos(T)-\sin(t)\sin(T)} \tag{1} \end{align} $$ if $T=2\pi$, then Eq.(1) becomes $$ \begin{align} f(x+T) &= \sqrt{\sin(x+T)}+ \sqrt{\cos(x+T)}\\ &= \sqrt{\sin(x)}+\sqrt{\cos(x)} \end{align} $$ this shows that the function is a periodic and $T=2\pi$,
A semi-decent way might be this. The general information about periodicity is given here.
We know that $\sin(x)$ is periodic with $2\pi$. If $f(x)$ is periodic with $a$, so is $g(f(x))$, hence $\sqrt{\sin(x)}$ is also periodic with $2\pi$. For the same reason, $\sqrt{\cos(x)}$ is also periodic with $2\pi$.
When two periodic functions are summed/multiplied the least common multiple of the individual functions is the new period. Therefore, $2\pi$ is a period of the $\sqrt{\sin(x)}+\sqrt{\cos(x)}$. But we do not yet know if it is the fundamental period.
Since it is the summation of two functions of the same period, $\pi$ might also be a period. (I don't have a good proof of this statement, this is why this is semi-decent, but you can think of addition of shifted impulse trains or frequency multiplier).
$$ \sqrt{\sin(x)}+\sqrt{\cos(x)}\stackrel{?}{=}\sqrt{\sin(x+\pi)}+\sqrt{\cos(x+\pi)}\\ \sqrt{\sin(x)}+\sqrt{\cos(x)}\stackrel{?}{=}\sqrt{-\sin(x)}+\sqrt{\cos(x)}, $$ such that $$ \sqrt{\sin(x)}+\sqrt{\cos(x)}\neq\sqrt{\sin(x+\pi)}+\sqrt{\cos(x+\pi)}, $$ hence $\pi$ is not a period. This leaves $2\pi$ as the fundamental period.