Science:Math Exam Resources/Courses/MATH110/December 2016/Question 07 (b)

MATH110 December 2016

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Question 07 (b)
Consider the function $f(x)={\frac {\sin(x)}{1-\cos(x)}}$. (b) Find all values of $x$ in the interval ${\textstyle [-2\pi ,2\pi ]}$ such that ${\textstyle f(x)=0}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Any solution must lie inside the domain.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. If ${\textstyle f(x)=0}$, then $${\dfrac {\sin(x)}{1-\cos(x)}}=0.$$ Since on the domain, the denominator doesn't vanish, it is equivalent with $$\sin(x)=0.$$ For ${\textstyle -2\pi \leq x\leq 2\pi }$, this happens when $$x=-2\pi ,-\pi ,0,\pi ,{\text{ or }}2\pi .$$ However, according to the Hint, ${\textstyle x}$ must not be an integer multiple of ${\textstyle 2\pi }$. Excluding ${\textstyle -2\pi ,0,2\pi }$, the remaining possible values are $$x=-\pi {\text{ or }}\pi .$$ Answer: If ${\textstyle f(x)=0}$ and ${\textstyle x}$ is in ${\textstyle [-2\pi ,2\pi ]}$, then ${\textstyle x={\color {blue}-\pi {\mbox{ or }}\pi }}$.