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Math Help - Trigonometric equations

  1. #1
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    Trigonometric equations

    Am I correct on the first one?
    A little help with second one please.

    \text{1.}) sin x + sin 3x = 0

    sin x + 3sin x - 4 sin^3 x = 0

    -4sin x(sin^2x - 1) = 0

    sin x = 0

    x = {\pi}n

    sin^2x - 1 = 0

    1sin x = 1

    x =  {\pi} + 2 {\pi}k

    2sinx = -1

    x = - \frac{\pi}{2} + 2{\pi}k

    x = - \frac{\pi}{2} + 2{\pi}m

    x(1) = 0

    x(2) = {\pi}

    x(3) = \frac{\pi}{2}

    0\le x \le{\pi}



    \text{2.})sin{\theta} + sin2{\theta} + sin3{\theta} = 0
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  2. #2
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    Quote Originally Posted by Hellbent View Post
    Am I correct on the first one?
    A little help with second one please.

    \text{1.}) sin x + sin 3x = 0

    sin x + 3sin x - 4 sin^3 x = 0

    -4sin x(sin^2x - 1) = 0

    sin x = 0

    x = {\pi}n

    sin^2x - 1 = 0 Correct up to here...
    \sin^2{x} = 1

    \sin{x} = \pm 1...

    So x = \left\{\frac{\pi}{2}, \frac{3\pi}{2}\right\} + 2\pi n

    x = \frac{\pi}{2} + \pi n.


    So putting it together you have

    x = \left\{0,  \frac{\pi}{2}\right\} + \pi n.
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  3. #3
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    Quote Originally Posted by Hellbent View Post
    \text{2.})sin{\theta} + sin2{\theta} + sin3{\theta} = 0
    \sin{\theta} + 2\sin{\theta}\cos{\theta} + 3\sin{\theta} - 4\sin^3{\theta} = 0

    4\sin{\theta} + 2\sin{\theta}\cos{\theta} - 4\sin^3{\theta} = 0

    \sin{\theta}(4 + 2\cos{\theta} - 4\sin^2{\theta}) = 0

    \sin{\theta}[4(1 - \sin^2{\theta}) + 2\cos{\theta}] = 0

    \sin{\theta}(4\cos^2{\theta} + 2\cos{\theta}) = 0

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


    Case 1: \sin{\theta} = 0

    \theta = \pi n.


    Case 2: \cos{\theta} = 0

    \theta = \frac{\pi}{2} + \pi n


    Case 3: 2\cos{\theta} + 1 = 0

    2\cos{\theta} = -1

    \cos{\theta} = -\frac{1}{2}

    \theta = \left\{ \pi - \frac{\pi}{3}, \pi + \frac{\pi}{3} \right\} + 2\pi n

    \theta = \left \{ \frac{2\pi}{3}, \frac{4\pi}{3} \right\} + 2\pi n.


    So putting it all together:

    \theta = \left \{ 0, \frac{\pi}{2}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{3\pi}{2} \right\} + 2\pi n.
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  4. #4
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    Ah! Thanks for your help
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  5. #5
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    having 2\sin\theta\cos\theta(2\cos\theta+1)=0, we have that this equals \sin(2\theta)(2\cos\theta+1)=0 so \theta=\frac{k\pi}2 and the other solutions already stated by Prove It.
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