Finding the fundamental period of $1+\frac{\cos x}{\sin 3x}$

Question

I apologize for the stupid question, but even though I know the technique for finding the period of trigonometric functions and sums of trig functions, I cannot figure out how to solve the following problem.

I need to find the fundamental period of the following:

$y(x)=1+\frac{\cos x}{\sin 3x}$

Clearly the period of $cos x$ is $2\pi$and the period of $\sin 3x$ is $2\pi/3$...

Wolfram Alpha says this is periodic of period pi. The techniques I think could work are to find a trig substitution that turns this into an addition problem, but I dont know which substitution to use. Otherwise, I really don't know what to do.

Thank you for your time.

Answer 1

First, $\pi$ is a period, because $$1+\frac{\cos(x+\pi)}{\sin(3(x+\pi))}=1+\frac{\cos(x+\pi)}{\sin(3x+3\pi)}=1+\frac{-\cos(x)}{-\sin(3x)}=1+\frac{\cos(x)}{\sin(3x)}.$$

Second, there can be no period smaller than $\pi$. To see this, consider where the function takes on the value $1$. It must be at the points $x$ where $\cos(x)=0$. This holds when $$x=\dots,-5\pi/2,-3\pi/2,-\pi/2,\pi/2, 3\pi/2, 5\pi/2,\dots$$

So the fundamental period is indeed $\pi$.

Answer 2

If $$f(x)=1+\dfrac{\cos x}{\sin3x},$$

$f(x+t)-f(x)$

$$=\dfrac{2\cos(x+t)\sin3x-2\cos x\sin3(x+t)}{2\sin3x\sin3(x+t)}$$

The numerator$$=\sin(4x+t)+\sin(2x-t)-\sin(4x+3t)-\sin(2x+3t)$$

$$=-2\sin t[\cos(4x+2t)+\cos(2x+t)]$$

So, we need $\sin t=0,t=?$