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Math Help - Inverse Trig Function Question.

  1. #1
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    Inverse Trig Function Question.

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

    My attempted solution:

    sin(ex/4) = -.5

    arcsin(-.5) = e(x/4)


    I understand why this is where I went wrong;
    because e(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
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  2. #2
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    Re: Inverse Trig Function Question.

    Quote Originally Posted by nicksbyman View Post
    I need to solve 2sin(e(x/4))+1=0 for x

    My attempted solution:

    sin(ex/4) = -.5

    arcsin(-.5) = e(x/4)


    I understand why this is where I went wrong;
    because e(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 2\pi, so

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