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Solve the given equation. Let k be any integer.

$$\csc^2 θ = 5 \cot θ + 7$$

I just need the first step or two please. I tried converting it:

$$\frac{1}{\sin^2 θ} = \frac {5\cosθ}{\sinθ} + 7$$

Then I tried a number of different ways to simplify it but it didn't work out

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    $\begingroup$ $\cot \theta = \frac{\cos \theta}{\sin \theta}$, not $\frac{\sin \theta}{\cos \theta}$. $\endgroup$ – MJD Aug 4 '15 at 18:11
  • $\begingroup$ @MJD thanks lol. give me a minute to try to solve it now $\endgroup$ – TheNewGuy Aug 4 '15 at 18:13
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HINT:

$$\csc^2 x=1+\cot^2 x$$

Then, solve a quadratic equation for $\cot x$

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  • $\begingroup$ thanks I forgot about that identity $\endgroup$ – TheNewGuy Aug 4 '15 at 18:18
  • $\begingroup$ @TheNewGuy You are more that welcome! My pleasure. $\endgroup$ – Mark Viola Aug 4 '15 at 18:19
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Notice, $$csc^2\theta=5\cot \theta+7$$

$$\cot^2\theta+1=5\cot \theta+7$$ $$\cot^2\theta-5\cot \theta-6=0$$ $$(\cot\theta -6)(\cot\theta+1)=0$$ $$\implies \cot\theta-6=0 \iff \tan \theta=\frac{1}{6}\iff \color{blue}{\theta=n\pi+\tan^{-1}\left(\frac{1}{6}\right)}$$

$$\implies \cot\theta+1=0 \iff \tan \theta=-1=-\tan\frac{\pi}{4} \iff \color{blue}{\theta=n\pi-\frac{\pi}{4}}$$ Where, $\color{blue}{n}$ is any integer

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$$\text{Recall: }\csc^2 \theta = 1 + \cot^2 \theta $$ So we have the following: $$\cot^2 \theta + 1 = 5\cot \theta + 7$$ $$\cot^2 \theta - 5\cot \theta - 6 = 0$$ $$\left(\cot\theta - 6\right)\left(\cot\theta + 1\right) = 0$$ $$\dots$$

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