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Math Help - Solving Trig Equations, need help.

  1. #1
    corsair29bd
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    Solving Trig Equations, need help.

    Im stuck on 2 problems:

    1. cos2x(2cos+1) (I dont know even where to start)

    and


    tan3x=1
    this one im beter off on

    i relize that tan=1 on (pie/4)

    diveded by three for the three rotations you get (pie/12)

    i know the answers are (pie/12 + pie N) the book says they are (pie/12, 9pie/12, 17pie/12, 5pie/12, 13pie/12, 21pie /12) how did it get these


    thanks
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by corsair29bd View Post
    Im stuck on 2 problems:
    1. cos2x(2cos+1) (I dont know even where to start)
    Is there an equation here to solve?

    First off, the symbol \pi is "pi" not "pie."

    cos(2x) = 2cos^2(x) - 1 so:

    cos(2x) \cdot (2cos(x)+1) = (2cos^2(x) - 1)(2cos(x)+1)

    = 4cos^3(x) + 2cos^2(x) - 2cos(x) - 1

    Now write your equation and we can help you further.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by corsair29bd View Post
    tan3x=1
    tan(3x) = 1

    3x = tan^{-1}(1) = \frac{\pi}{4}

    Now, this is a reference angle. So actually we have the 2 equations:
    3x = \frac{\pi}{4}

    3x = \frac{5\pi}{4}

    (We ignore quadrants II and IV since tangent is negative there.)

    So x = \frac{\pi}{12} and x = \frac{5 \pi}{12}.

    Now, we are dividing by 3 so there should be 3 solutions associated with each x solution we have. (This is because we can add multiples of 2\pi to each solution for 3x.)

    x = \frac{\pi}{12}, \frac{\pi}{12} + \frac{2\pi}{3}, \frac{\pi}{12} + \frac{4\pi}{3}

    x = \frac{5\pi}{12}, \frac{5\pi}{12} + \frac{2\pi}{3}, \frac{5\pi}{12} + \frac{4\pi}{3}

    And this generates the list you gave as a solution.

    -Dan
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