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

Question
Functions asked 2021-01-23
$$x^{4}(x^{2}-3)^{6}\sin^{2}8x$$

## Answers (1) 2021-01-24
If the question is to find the zeros of the function, then here is the answer.
We have $$x^{4}(x^{2}−3)^{6}\sin^{2}8x=0$$ only when
$$x^{4}=0, or, x^{2}-3)^{6}=0, or, sin^{2}8x=0$$
This is equivalent to
$$x=0, or, x^{2}-3 = 0, or, \sin8x=0$$
We know that sine function is zero for angles nπ,nπ, where nn is an integer. Therefore, we get
$$x=0, or, x=\pm\sqrt3, or, x = \frac{8}{n\pi}.$$
If you need more explanation, or wanted something else then feel free to leave a comment. I will update the answer accordingly. ​

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