Inverse Trig Function Question.

I need to solve 2sin(eˆ(x/4))+1=0 for x

My attempted solution:

sin(eˆx/4) = -.5

arcsin(-.5) = eˆ(x/4)

ˆ

I understand why this is where I went wrong;

because eˆ(x/4) is never negative, but I don't understand how I would solve this equation, and more importantly I don't understand why I couldn't use inverse trig functions to solve this problem. **I thought that if sin x = y then arcsiny = x**

Re: Inverse Trig Function Question.

Quote:

Originally Posted by

**nicksbyman** I need to solve 2sin(eˆ(x/4))+1=0 for x

My attempted solution:

sin(eˆx/4) = -.5

arcsin(-.5) = eˆ(x/4)

ˆ

I understand why this is where I went wrong;

because eˆ(x/4) is never negative, but I don't understand how I would solve this equation, but more importantly I don't understand why I couldn't use inverse trig functions to solve this problem.

You need to realise that the sine function is negative in the third and fourth quadrants, and has a period of $\displaystyle \displaystyle 2\pi$, so

$\displaystyle \displaystyle \begin{align*} \sin{\left(e^{\frac{x}{4}}\right)} &= -\frac{1}{2} \\ e^{\frac{x}{4}} &= \left\{\pi + \arcsin{\left(\frac{1}{2}\right)}, 2\pi - \arcsin{\left(\frac{1}{2}\right)} \right\} + 2\pi n \textrm{ where }n \in \mathbf{Z^{+} \cup \{0\}} \end{align*}$

We choose $\displaystyle \displaystyle n$ to only represent nonnegative integers, because as you said, the exponential function is always positive. Go from here.