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So I asked this before with a similar question, and while I got the answer, I still don't understand how to figure out what integer(s) $k$ is.

An equation is give (express your answer in terms of k, where k is any integer)

$$3 csc^2 θ = 4$$

(a) Find all solutions of the equation.

(b) Find the solutions in the interval $[0, 2π)$.

(a) I've already gotten part A correct (I boxed the answer):

I still don't know what integer(s) to plug in for $k$ or how to find out what those integer(s) are. I need to know how to find $k$ so I can yield results between the given interval.

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3 Answers 3

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First of all, $\theta=\frac{\pi}{3}+2k\pi$ and $\theta=\frac{4\pi}{3}+2k\pi$ can be written as $\theta=\frac{\pi}{3}+k\pi$.

Similarly, $\theta=-\frac{\pi}{3}+2k\pi$ and $\theta=\frac{2\pi}{3}+2k\pi$ can be written as $\theta=\frac{2\pi}{3}+k\pi$.

To know what integers to plug in for $k$, you can solve the following inequality for $k$ : $$0\le \theta=\frac{\pi}{3}+k\pi\lt 2\pi$$ $$-\frac{\pi}{3}\le \pi k\lt 2\pi-\frac{\pi}{3}$$ $$-\frac 13=\frac{-\frac{\pi}{3}}{\pi}\le k\lt \frac{2\pi-\frac{\pi}{3}}{\pi}=\frac 53$$ So, in this case, $k=0,1$. So, $\theta=\frac{\pi}{3}+\pi\cdot 0=\frac{\pi}{3}$ and $\theta=\frac{\pi}{3}+1\cdot \pi=\frac{4}{3}\pi$.

On the other hand, $$0\le \theta=\frac{2\pi}{3}+k\pi\lt 2\pi$$ $$-\frac{2\pi}{3}\le \pi k\lt 2\pi-\frac{2\pi}{3}$$ $$-\frac 23=\frac{-\frac{2\pi}{3}}{\pi}\le k\lt \frac{2\pi-\frac{2\pi}{3}}{\pi}=\frac 43$$ So, in this case, $k=0,1$. So, $\theta=\frac{2\pi}{3}+\pi\cdot 0=\frac{2\pi}{3}$ and $\theta=\frac{2\pi}{3}+1\cdot \pi=\frac{5}{3}\pi$.

Thus, the answer is $$\color{red}{\theta=\frac{1}{3}\pi,\quad\frac{2}{3}\pi,\quad\frac{4}{3}\pi,\quad\frac{5}{3}\pi}.$$

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$$\csc^2\theta =\frac{4}{3}$$

$$\csc\theta =\frac{2}{\sqrt3}\text{ or }\csc\theta =-\frac{2}{\sqrt3}$$

$$\sin\theta =\frac{\sqrt3}{2}\text{ or }\sin\theta =-\frac{\sqrt3}{2}$$

$$\boxed{\color{blue}{\theta=\pi n-\frac{4 \pi}{3}}}$$

$$\boxed{\color{blue}{\theta=\pi n-\frac{2 \pi}{3}}}$$

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  • $\begingroup$ Yup, I already got that part.... $\endgroup$
    – TheNewGuy
    Aug 4, 2015 at 22:26
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Alright, let's go over the process again, we have our $\theta$'s, $ \theta = \pi n \pm \frac{4\pi}{3} , \pi n \pm \frac{2\pi}{3}$. Those are our general solutions, which you seem to have no trouble finding, if you want to find all solutions in an interval (as discussed by Mnifldz last time) we are going to plug in values of $n$ up to, but not including, those values which would push us into the region reaching or surpassing $2\pi$.

For $n=0, \theta = \frac{4\pi}{3}, \frac{2\pi}{3}$.

For $n=1, \theta = \frac{\pi}{3}, \frac{5\pi}{3}$.

$n\ge 2$ would not work since then we'd be outside of our acceptable interval as the problem stated, thus we've found all solutions in the interval.

Note: We exclude the negative values since they too are outside of the given interval, as well as $\frac{7\pi}{3}$.

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