# Question 7392e

Mar 27, 2017

$3 \csc x - 2 \sqrt{3} = 0$

Add $2 \sqrt{3}$ to both sides
$3 \csc x \cancel{- 2 \sqrt{3}} \cancel{+ 2 \sqrt{3}} = \cancel{0 +} 2 \sqrt{3}$
$3 \csc x = 2 \sqrt{3}$

Divide both sides by $3$
$\frac{\cancel{3} \csc x}{\cancel{3}} = \frac{2 \sqrt{3}}{3}$
$\csc x = \frac{2 \sqrt{3}}{3}$

Replace $\csc x$ with $\frac{1}{\sin} x$ and $\frac{\sqrt{3}}{3}$ with $\frac{1}{\sqrt{3}}$
$\frac{1}{\sin} x = \frac{2}{\sqrt{3}}$

Raise both sides to the power of $\text{-} 1$ and simplify
${\left(\frac{1}{\sin} x\right)}^{\text{-"1)=(2/sqrt3)^("-} 1}$
$\sin x = \frac{\sqrt{3}}{2}$

Use knowledge of special angles (or a calculator if you don't need an exact answer)
$x = \frac{\pi}{3}$

Realize that $\sin \left(\pi - \theta\right) = \sin \left(\theta\right)$
$x = \frac{\pi}{3} , \frac{2 \pi}{3}$

Since $x$ is unbounded and $f \left(x + 2 n \pi\right) = f \left(x\right)$ when $f \left(\theta\right)$ is a trig function and $n \in \mathbb{Z}$

{x:x=pi/3+2npi,(2pi)/3+2npi;ninZZ}#