# How do you solve sqrt(3) - 2Sin(6x) = 0 from [0,2pi]?

Jul 24, 2015

$\sqrt{3} - 2 \sin \left(6 x\right) = 0$ with $x \epsilon \left[0 , 2 \pi\right]$
$\rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{\pi}{9} \mathmr{and} \frac{\pi}{18}$

#### Explanation:

Temporarily replace $6 x$ with $\theta$

$\sqrt{3} - 2 \sin \left(\theta\right) = 0$

$\rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$$\sin \left(\theta\right) = \frac{\sqrt{3}}{2}$

within the range $\left[0 , 2 \pi\right]$ this is one of the standard angles with
$\textcolor{w h i t e}{\text{XXXX}}$$\theta = \frac{\pi}{3}$ or $\theta = \frac{2 \pi}{3}$

Since $\theta = 6 x$
$\textcolor{w h i t e}{\text{XXXX}}$$6 x = \frac{\pi}{3}$ or $6 x = \frac{2 \pi}{3}$

$\rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{\pi}{18}$ or $x = \frac{\pi}{9}$