Question

# x^{4}(x^{2}-3)^{6}sin^{2}8x

Functions
$$\displaystyle{x}^{{{4}}}{\left({x}^{{{2}}}-{3}\right)}^{{{6}}}{{\sin}^{{{2}}}{8}}{x}$$

2021-01-07
If the question is to find the zeros of the function, then here is the answer.
We have $$\displaystyle{x}^{{{4}}}{\left({x}^{{{2}}}−{3}\right)}^{{{6}}}{{\sin}^{{{2}}}{8}}{x}={0}$$ only when
$$\displaystyle{x}^{{{4}}}={0},{\quad\text{or}\quad},{x}^{{{2}}}-{3}{)}^{{{6}}}={0},{\quad\text{or}\quad},{{\sin}^{{{2}}}{8}}{x}={0}$$
This is equivalent to
$$\displaystyle{x}={0},{\quad\text{or}\quad},{x}^{{{2}}}-{3}={0},{\quad\text{or}\quad},{\sin{{8}}}{x}={0}$$
We know that sine function is zero for angles nπ,nπ, where nn is an integer. Therefore, we get
$$\displaystyle{x}={0},{\quad\text{or}\quad},{x}=\pm\sqrt{{3}},{\quad\text{or}\quad},{x}={\frac{{{8}}}{{{n}\pi}}}.$$
If you need more explanation, or wanted something else then feel free to leave a comment. I will update the answer accordingly. ​