If $g(x)=\frac1{1-2\sin^2x}$ and $f(x)=\sin 2x$ , what is the period of $\frac{f(x)}{g(x)}$?

Question

If $g(x)=\dfrac1{1-2\sin^2x}$ and $f(x)=\sin 2x$ , what is the period of the function $\dfrac{f(x)}{g(x)}$ ?

We have $$\dfrac{f}{g}=\sin2x(1-2\sin^2x)=\sin(2x)\cos(2x)=\frac12\sin(4x)$$ Where $\cos2x\neq0$ or $x\neq\frac{k\pi}2+\frac{\pi}4$.

I know the period of $\frac12\sin(4x)$ is $\frac{\pi}2$. But does $x\neq\frac{k\pi}2+\frac{\pi}4$ affect on the period?

Answer 1

$$\frac{\sin 2x}{1-2\sin^2 x}=\frac{\sin 2x}{\cos 2x}=\tan 2x$$ Since $\tan x$ has a period of $\pi$, then $\tan 2x$ will have a period of $\pi/2$ But since we have a restriction on the denominator, the period will be $\pi$