# How do you solve tanx+sqrt3=0?

Jun 16, 2015

$\tan \left(x\right) + \sqrt{3} = 0$ has two solutions:

${x}_{1} = - \frac{\pi}{3}$

${x}_{2} = \pi - \frac{\pi}{3} = \frac{2 \pi}{3}$

#### Explanation:

The equation $\tan \left(x\right) + \sqrt{3} = 0$ can be rewritten as
$\tan \left(x\right) = - \sqrt{3}$

Knowing that $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$

and knowing some specific values of $\cos$ and $\sin$ functions:

$\cos \left(0\right) = 1$ ; $\sin \left(0\right) = 0$

$\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ ; $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

$\cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ ; $\sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

$\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$ ; $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

$\cos \left(\frac{\pi}{2}\right) = 0$ ; $\sin \left(\frac{\pi}{2}\right) = 1$

as well as the following $\cos$ and $\sin$ properties:

$\cos \left(- x\right) = \cos \left(x\right)$ ; $\sin \left(- x\right) = - \sin \left(x\right)$
$\cos \left(x + \pi\right) = - \cos \left(x\right)$ ; $\sin \left(x + \pi\right) = - \sin \left(x\right)$

We find two solutions:

1) $\tan \left(- \frac{\pi}{3}\right) = \sin \frac{- \frac{\pi}{3}}{\cos} \left(- \frac{\pi}{3}\right) = \frac{- \sin \left(\frac{\pi}{3}\right)}{\cos} \left(\frac{\pi}{3}\right) = - \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = - \sqrt{3}$

2) $\tan \left(\pi - \frac{\pi}{3}\right) = \sin \frac{\pi - \frac{\pi}{3}}{\cos} \left(\pi - \frac{\pi}{3}\right) = \frac{- \sin \left(- \frac{\pi}{3}\right)}{- \cos \left(- \frac{\pi}{3}\right)} = \sin \frac{\frac{\pi}{3}}{- \cos \left(\frac{\pi}{3}\right)} = - \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = - \sqrt{3}$