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Q: Determine the number of solutions for $((\sin x)^2)(\csc x + 1) = 0$ over the interval $0 \leq x < 2\pi$ with the correct reasoning.

Correct answer: There is one solution because the solutions of $(\sin x)^2 = 0$ have to be rejected and $\csc x + 1 = 0$ has one solution in given interval.

My question: Why should $(\sin x)^2 = 0$ be rejected??

What I think should be the answer: There are three solutions because $(\sin x)^2 = 0$ has two solutions and $\csc x + 1 = 0$ has one solution in the given interval.

Thanks in advance!!

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  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. Also, when you read the responses, keep in mind that the notation $\sin^2x$ is often used as a shorthand for $(\sin x)^2$. $\endgroup$ Jun 24, 2016 at 10:10
  • $\begingroup$ Just like $x^2/x$ is not defined at $0$ despited being simplifiable to $x$ for $x\ne0$, so is your function undefined at $k\pi$ despite being simplifiable to $\sin x(1+\sin x)$ for all the other $x$. $\endgroup$ Jun 24, 2016 at 10:15
  • $\begingroup$ Maybe easier to analyse if we express $((\sin x)^2)(\csc x + 1)$ as $\sin x (1+\sin x)$ $\endgroup$
    – John Joy
    Jun 24, 2016 at 21:16

2 Answers 2

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$\sin x=0$ is rejected because when $\sin x=0$, the value $\csc x$ does not exist, so $(\sin^2 x)(1+\csc x)$ does not exist, so it can't be equal to $0$ or to anything else either.

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Let's say $(\sin x)^2=0$. Then, take square root of both sides to get $\sin x=0$. Now, we have either $x=0$ to $x=\pi$. Then, substitute back in: $$x=0 \implies (\sin 0)^2(\csc 0+1)=\text{undefined because} \csc 0 \text{ is undefined}$$ $$x=\pi \implies (\sin \pi)^2(\csc \pi+1)=\text{undefined because} \csc \pi \text{ is undefined}$$ Always remember to substitute solutions back into the original equation to make sure you do not have any undefined, like $\csc 0$ or $\frac{1}{0}$, or any extraneous solutions.

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    $\begingroup$ I'd say that this is the right answer (although the way it's given it is rather unclear), although it's worth noting that while the function is technically undefined at the points where $\sin^2 x = 0$, they are removable singularities (i.e. you can specify a value for the function at those points in a way that keeps the function's continuity), and if that's the case then at those points the function is zero. $\endgroup$
    – ConMan
    Jun 24, 2016 at 1:31
  • $\begingroup$ @ConMan I think my answer is more clear now. However, I don't know what a "singularity" is (I have not learned complex analysis yet), so I can't write an answer about that. $\endgroup$ Jun 24, 2016 at 1:41
  • $\begingroup$ A singularity is just a "difficult" point or region. Sometimes it's removable i.e. you can put a value in and the function still works), sometimes it's more fundamental (e.g. 1/x when x = 0). $\endgroup$
    – ConMan
    Jun 24, 2016 at 6:03

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