Prove that $f(x)=3\sin⁡(3\pi \ x/4)\ + \ 2 \cos(\pi \ x/3)$ is neither odd or even function

Question

I know $f(x)$ is neither even or odd if $f (-x) \ne -f (x)$ and $f (-x) \ne f (x)$

This is what I have done so far. $$f(x)=3 \sin⁡(\pi x/4)+2 \cos⁡(\pi x/3)$$ $$\begin{align} f(-x)&=3 \sin⁡(-3πx/4)+2 \cos⁡(-πx/3) \\ &=-3 \sin⁡(3πx/4)+2 \cos⁡(-\pi x/3)\\ \end{align}$$

$$f (-x)\ne f (x),\quad f(-x) \ne f (x)$$

Is the solution sufficient for a full proof?

Thank you in advance!

Answer 1

You have the correct idea, but you can improve on the way to present it.

$f(-x)=-3\sin(3\pi x/4)+2\cos(\pi x /3)$

On the one hand, $g(x)=f(-x)+f(x)=4\cos(\pi x /3)$, and since $g(0) \neq 0$ we have $g(x) \neq 0$ and therefore f is not odd.

On the other hand, $h(x)=f(x)-f(-x)=6\sin(3\pi x/4)$, and since $h(1) \neq 0$ we have $h(x) \neq 0$ and therefore f is not even.

Answer 2

To prove it, you need only one counterexample, and the simple case $x=\pm1$ suffices.