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Math Help - [Pre-Calc] Finding all solutions of a trig equation

  1. #1
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    [Pre-Calc] Finding all solutions of a trig equation

    sin(2x) - sinx = 2cos^2 (x) - cosx


    Can someone show me the steps on how to find all the solutions in the interval [0,2pi)?
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  2. #2
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    Re: [Pre-Calc] Finding all solutions of a trig equation

    Quote Originally Posted by albo23 View Post
    sin(2x) - sinx = 2cos^2 (x) - cosx


    Can someone show me the steps on how to find all the solutions in the interval [0,2pi)?
    2\sin{x}\cos{x} - \sin{x} = 2\cos^2{x} - \cos{x}

    \sin{x}(2\cos{x} - 1) = \cos{x}(2\cos{x} - 1)

    edit ... pulled out the factor incorrectly.

    $#it happens.
    Last edited by skeeter; April 25th 2012 at 01:56 PM. Reason: fixed the botched factoring job
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  3. #3
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    Re: [Pre-Calc] Finding all solutions of a trig equation

    Quote Originally Posted by albo23 View Post
    sin(2x) - sinx = 2cos^2 (x) - cosx


    Can someone show me the steps on how to find all the solutions in the interval [0,2pi)?
    Hi albo23,

    Slightly different approach. Use the double angle identity \sin 2x = 2 \sin x \cos x to replace \sin 2x

    Then,

    2 \sin x \cos x - \sin x = 2 \cos^2 x - \cos x

    2 \sin x \cos x - \sin x - 2 \cos^2 x + \cos x = 0

    \sin x (2 \cos x - 1) - \cos x(2 \cos x - 1)=0

    (\sin x - \cos x)(2 \cos x - 1)=0

    \sin x - \cos x = 0 \text{ or } 2 \cos x -1 = 0

    In which case,

    \sin x = \cos x \text{ or } \cos x = \frac{1}{2}

    In the first case, \sin x = \cos x \text{ at } \boxed{\frac{\pi}{4}+2\pi k \text{ and at }\frac{5\pi}{4}+2\pi k}

    In the second case, \cos x = \frac{1}{2} \text{ at } \boxed{\frac{\pi}{3}+2\pi k \text{ and at }\frac{5\pi}{3}+2\pi k}

    My answers differ from Skeeter's. Who is correct?




    Last edited by masters; April 25th 2012 at 12:00 PM.
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    Re: [Pre-Calc] Finding all solutions of a trig equation

    Quote Originally Posted by masters View Post
    My answers differ from Skeeter's. Who is correct?
    Skeeter got a bit carried away.
    \color{red}{\cos{x}(2\sin{x} - 1)} - \cos{x}(2\cos{x} - 1) = 0
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