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Math Help - Please help with equation involving inverse trigonometric functions

  1. #1
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    Please help with equation involving inverse trigonometric functions

    Solve for x:
    arccos x +2arcsin(√3/2)= π/3

    I keep getting 1/2 for my answer but it does not check. Here's my process

    arccos x + 2arcsin(√3/2)= π/3

    arccos x = π/3 - 2arcsin(√3/2)

    x=cos[π/3 - 2arcsin(√3/2)]

    Using cosine of a difference identity:
    x=cos(π/3)cos2[arcsin(√3/2)] + sin(π/3)sin2[arcsin(√3/2)]

    x=(1/2)cos(2π/3) + (√3/2)sin(2π/3)

    x=(1/2)(-1/2) + (√3/2)(√3/2)

    x= (-1/4) + (3/4)

    x=2/4=1/2

    Could someone please tell me what I'm doing wrong?
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  2. #2
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    Re: Please help with equation involving inverse trigonometric functions

    arccos x +2arcsin(√3/2)= π/3
    \arccos{x} + 2 \cdot \frac{\pi}{3} = \frac{\pi}{3}

    \arccos{x} = -\frac{\pi}{3}

    the range of the \arccos function is [0,\pi] ... this equation has no solution.
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