# How do you find the domain of g(x) = (3^(x+1)) / (sin x + cos x) ?

Feb 17, 2017

$\left\{x | x \ne \frac{3 \pi}{4} + \pi n , x \in \mathbb{R} , \text{where } n \in \mathbb{Z}\right\}$

#### Explanation:

For this problem, we have to ask ourselves a couple of questions.

•When does $\sin x + \cos x$ equal $0$.
•What is the domain of ${3}^{x + 1}$

I'll start by answering the first one. Solve the trigonometric equation.

$\sin x + \cos x = 0$

Square both sides

${\left(\sin x + \cos x\right)}^{2} = {0}^{2}$

${\sin}^{2} x + 2 \sin x \cos x + {\cos}^{2} x = 0$

Apply ${\sin}^{2} x + {\cos}^{2} x = 1$:

$1 + 2 \sin x \cos x = 0$

Use $2 \sin x \cos x = \sin 2 x$:

$1 + \sin 2 x = 0$

$\sin 2 x = - 1$

$2 x = \arcsin \left(- 1\right)$

$2 x = \frac{3 \pi}{2} + 2 \pi n$ because the sine function has a period of $2 \pi$

$x = \frac{3 \pi}{4} + \pi n$

This means that whenever $x = \frac{3 \pi}{4} + \pi n$, $n$ an integer, the graph of $g \left(x\right) = {3}^{x + 1} / \left(\sin x + \cos x\right)$ will have vertical asymptotes.

${3}^{x + 1}$ is your run of the mill exponential function ; it will have a domain of all the real numbers, but will have a restricted range (we aren't dealing with range in this problem, though, so I won't go into detail there).
This means that the domain of $g \left(x\right)$ is $\left\{x | x \ne \frac{3 \pi}{4} + \pi n , x \in \mathbb{R} , \text{where } n \in \mathbb{Z}\right\}$.