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 , so

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