# How do I solve 3 sin x+2 tan x=0 for 0<=x<=360 ?

Sep 28, 2015

$x = 0$,
$x = 180$,
$x = 360$,
$x = \arccos \left(- \frac{2}{3}\right)$,
$x = - \arccos \left(- \frac{2}{3}\right)$

#### Explanation:

The first step is to determine the domain of the functions involved. With the restriction $0 \le x \le 360$ (degrees, I assume) we have exclude the points where $\tan \left(x\right)$ is not defined.
By definition, $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$
Therefore, we have to exclude points where $\cos \left(x\right) = 0$, that is the domain is described as
$x \ne 90$ and $x \ne 270$

With the above restrictions in mind we can transform the equation as follows:
$3 \sin \left(x\right) + 2 \sin \frac{x}{\cos} \left(x\right) = 0$

The immediate temptation to reduce the above equation by $\sin \left(x\right)$ should be accompanied by checking if it can be equal to zero to avoid losing solutions.
Indeed, $\sin \left(x\right) = 0$ can occur within a domain defined above.
It happens when
$x = 0$, $x = 180$ and $x = 360$.
So the three values above are solutions.

After specifying these three solutions we can reduce the equation by $\sin \left(x\right)$ getting
$3 + \frac{2}{\cos} \left(x\right) = 0$,
from which follows
$\cos \left(x\right) = - \frac{2}{3}$,
solutions of this are
$x = \arccos \left(- \frac{2}{3}\right)$ and $x = - \arccos \left(- \frac{2}{3}\right)$

CHECKING

The first three solutions ($0$, $180$ and $360$) cause both $\sin \left(x\right)$ and $\tan \left(x\right)$ to be equal to $0$, therefore the left side will be equal to $0$.

The next two values ($\arccos \left(- \frac{2}{3}\right)$ and $- \arccos \left(- \frac{2}{3}\right)$ result in the following:

$3 \sin \left(\arccos \left(- \frac{2}{3}\right)\right) + 2 \sin \frac{\arccos \left(- \frac{2}{3}\right)}{\cos} \left(\arccos \left(- \frac{2}{3}\right)\right) =$

$= \sin \left(\arccos \left(- \frac{2}{3}\right)\right) \cdot \left[3 + \frac{2}{- \frac{2}{3}}\right] =$

$= \sin \left(\arccos \left(- \frac{2}{3}\right)\right) \cdot \left[3 - 3\right] = 0$

$3 \sin \left(- \arccos \left(- \frac{2}{3}\right)\right) + 2 \sin \frac{- \arccos \left(- \frac{2}{3}\right)}{\cos} \left(- \arccos \left(- \frac{2}{3}\right)\right) =$

$= \sin \left(- \arccos \left(- \frac{2}{3}\right)\right) \cdot \left[3 + \frac{2}{- \frac{2}{3}}\right] =$

$= \sin \left(- \arccos \left(- \frac{2}{3}\right)\right) \cdot \left[3 - 3\right] = 0$

All is checked.