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Math Help - Solving a trig equation - Is my work correct?

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    Solving a trig equation - Is my work correct?

    Find all solutions of the equation in the interval [0, 2pi):
    sin (x + pi/6) - sin (x - pi/6) = 1/2

    My work:
    sin x cos pi/6 + cos x sin pi/6 - sin x cos pi/6 - cos x sin pi/6 = 1/2
    2 sin x cos pi/6 = 1/2
    (2 sin x) (sqrt3/2) = 1/2
    Now this is the part where I am confused,
    2 sin x = (1/2) / (sqrt3/2)
    2 sin x = (2) / (2 sqrt3)
    2 sin x = 1/sqrt 3
    sin x = sqrt3/2
    x = pi/3, 2pi/3

    Is this right?
    Any help is appreciated
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  2. #2
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    Quote Originally Posted by iluvmathbutitshard View Post
    Find all solutions of the equation in the interval [0, 2pi):
    sin (x + pi/6) - sin (x - pi/6) = 1/2

    My work:
    sin x cos pi/6 + cos x sin pi/6 - sin x cos pi/6 - cos x sin pi/6 = 1/2
    2 sin x cos pi/6 = 1/2
    (2 sin x) (sqrt3/2) = 1/2
    Now this is the part where I am confused,
    2 sin x = (1/2) / (sqrt3/2)
    2 sin x = (2) / (2 sqrt3)
    2 sin x = 1/sqrt 3
    sin x = sqrt3/2
    x = pi/3, 2pi/3

    Is this right?
    Any help is appreciated
    Hi iluvmathbutitshard,

    I think you have a sign wrong in your first step:

    \sin x \cos \frac{\pi}{6}+\cos x \sin \frac{\pi}{6}-(\sin x \cos \frac{\pi}{6}-\cos x \sin \frac{\pi}{6})=\frac{1}{2}

    \sin x \cos \frac{\pi}{6}+\cos x \sin \frac{\pi}{6}-\sin x \cos \frac{\pi}{6}+\cos x \sin \frac{\pi}{6}=\frac{1}{2}

    \frac{\sqrt{3}}{2}\sin x+\frac{1}{2} \cos x -\frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x=\frac{1}{2}


    \cos x = \frac{1}{2}

    x=\left\{\frac{\pi}{3}, \frac{5\pi}{3}\right\}
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