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Math Help - Solving equation for x

  1. #1
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    Solving equation for x

    1. arcsin(3x -pi) = 1/2

    2. arcsin rad(2x) = arccos rad(x)
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  2. #2
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    Re: Solving equation for x

    Two things:

    1) The argument for arcsin(x) and arccos(x) requires -1 \le x \le 1

    2) What does "rad(x)" mean?
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  3. #3
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    Re: Solving equation for x

    Quote Originally Posted by TKHunny View Post
    Two things:

    1) The argument for arcsin(x) and arccos(x) requires -1 \le x \le 1

    2) What does "rad(x)" mean?
    1) Correct...I still don't understand it though. How can an arcsin form an angle of 1/2?

    2) I posted the equations again below (in attachment). Little better format.
    Attached Thumbnails Attached Thumbnails Solving equation for x-capture.png  
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  4. #4
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    Re: Solving equation for x

    Quote Originally Posted by PleaseHelp View Post
    1) Correct...I still don't understand it though. How can an arcsin form an angle of 1/2?

    1/2 is an angle in radians
    1. as mentioned earlier, note that ...

    -1 \le 3x-\pi \le 1

    therefore the value of x is restricted to ...

    \frac{\pi - 1}{3} \le x \le \frac{\pi + 1}{3}

    solving the equation ...

    3x - \pi = \sin\left(\frac{1}{2}\right)

    x = \frac{1}{3}\left[\sin\left(\frac{1}{2}\right) + \pi\right] \approx 1.207

    2. since arcsin(something) = arccos(something else) , the angle these two expressions equal can only be in quad I (why is that?)

    let \theta = \arcsin{\sqrt{2x}}

    \cos{\theta} = \sqrt{1-2x}

    since the angles are equal ...

    \theta = \arccos{\sqrt{x}}

    \cos{\theta} = \sqrt{x}

    so ...

    \sqrt{1-2x} = \sqrt{x}

    solve for x
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