# Solve the trigonometric equation $\csc^2 \theta= 5 \cot \theta + 7$

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

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

HINT:

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

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

• thanks I forgot about that identity – TheNewGuy Aug 4 '15 at 18:18
• @TheNewGuy You are more that welcome! My pleasure. – Mark Viola Aug 4 '15 at 18:19

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

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