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Math Help - Finding all solutions of to the equation

  1. #1
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    Finding all solutions of to the equation

    Find all solutions to the equation in the interval [0, 2pi)

    Problem: sin 2x = -sin x

    My attempt:
    I'm thinking of making it equal to 0

    sin2x - sin x = 0
    factor out sin

    sin(2x-1) = 0
    2x-1= 0
    x =1/2

    I dont know if i'm going the right direction .

    sin(1/2) ?
    on unit circle sin 1/2 = pi/6, 5pi/6, 11pi/6
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  2. #2
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    Quote Originally Posted by yeunju View Post
    Find all solutions to the equation in the interval [0, 2pi)

    Problem: sin 2x = -sin x

    My attempt:
    I'm thinking of making it equal to 0

    sin2x - sin x = 0
    factor out sin

    sin(2x-1) = 0
    2x-1= 0
    x =1/2

    I dont know if i'm going the right direction .

    sin(1/2) ?
    on unit circle sin 1/2 = pi/6, 5pi/6, 11pi/6
    sorry to tell you this, but you are way off ...

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

    \sin(2x) + \sin{x} = 0

    use the double angle identity for sine ...

    2\sin{x}\cos{x} + \sin{x} = 0

    factor ...

    \sin{x}(2\cos{x} + 1) = 0

    set each factor equal to 0 ...

    \sin{x} = 0 ... x = 0 \, , x = \pi<br />

    \cos{x} = -\frac{1}{2} ... x = \frac{2\pi}{3} \, x = \frac{4\pi}{3}
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