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Math Help - Trig: Equations with Inverse Trigonometric Functions?

  1. #1
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    Unhappy Trig: Equations with Inverse Trigonometric Functions?

    I need help with solving this problem:

    arccos x + 2 arcsin (√3/2) = pi/3

    I keep getting 1/2, and when I plug it back into the equation it equals pi instead of pi/3. Here's what I did:


    arccos x + 2 arcsin (√3/2) = pi/3

    arccos x = pi/3 - 2 arcsin (√3/2)

    arccos x = -pi/3

    x = cos (-pi/3)

    x = 1/2


    What did I do wrong? What is the correct answer?

    I really need help now cuz my math test is Monday! D:
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    looks to me that you did everything correct.

    the range of y = \arccos{x} is 0 \leq y \leq \pi

    as I see it, this equation has no solution.
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  3. #3
    Senior Member chella182's Avatar
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    I would've done exactly the same as you did there . Odd.
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