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Math Help - [SOLVED] Another Cos Question. Urgent!!

  1. #1
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    [SOLVED] Another Cos Question. Urgent!!

    Barron's Math IIC SAT Subject Test Page 82

    Which of the following is a solution of cos3x=0.5

    a) 60 degrees

    b) (5pi)/3

    c)cos^-1 (1/6)

    d) cos^-1 ( \frac{\sqrt3}{2}

    e) \frac{1}{3}cos^-1(1/2)
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by fabxx View Post
    Barron's Math IIC SAT Subject Test Page 82

    Which of the following is a solution of cos3x=0.5

    a) 60 degrees

    b) \frac{5\pi}{3}

    c)cos^-1 (1/6)

    d) cos^-1 ( \frac{\sqrt3}{2}

    e) \frac{1}{3}cos^-1(1/2)
    Solve for x:

    \cos(3x)=\tfrac{1}{2}\implies \cos^{-1}\left[\cos(3x)\right]=\cos^{-1}\left(\tfrac{1}{2}\right)\implies 3x=\cos^{-1}\left(\tfrac{1}{2}\right)

    I'm sure the answer is pretty obvious now

    --Chris
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  3. #3
    Senior Member vincisonfire's Avatar
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     cos(3x) = \frac{1}{2}
     arcos(cos(3x)) = arccos(\frac{1}{2})
     3x = \frac{\pm\pi}{3}
     x = \frac{\pm\pi}{9}
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    Yeah it does thanks but when do you know to use arc-cos or sin? And, if I approach these kind of problems later, do i always use arc-cos?
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    Senior Member vincisonfire's Avatar
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    If x is within a cos, you use arccos to get it out of it.
    Same thing for sin tan or any trigo function.
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by fabxx View Post
    Yeah it does thanks but when do you know to use arc-cos or sin? And, if I approach these kind of problems later, do i always use arc-cos?
    It depends on the problem.

    If you had \sin (x+3)=\tfrac{\sqrt{3}}{2}, and you wanted to solve for x, you would apply arcsin to both sides, since we know that \sin^{-1}\left(\sin u\right)=u

    In your problem, you would have to apply arccos to both sides, since we know that \cos^{-1}\left(\cos u\right)=u

    Does this clarify things?

    --Chris
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