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Math Help - Help solving trigometric equations!

  1. #1
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    Help solving trigometric equations!

    1. 4cos^2 3x = 1
    2. cos(2x-pi/2)=root2/2
    3. cos2x=sinx+1
    4. sin6xcosx-cos6xsinx= - root3/2
    4. cos2x-4=3cosx-4sin^2x

    Please explain to me how to solve these!!

    Also, how can i type the root or squares out like I see people do in this forum? or is it an image?
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  2. #2
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    Quote Originally Posted by dorkymichelle View Post

    Also, how can i type the root or squares out like I see people do in this forum? or is it an image?
    http://www.mathhelpforum.com/math-he...-tutorial.html


    Please post one question per thread with your attempts at each question
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  3. #3
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    Quote Originally Posted by dorkymichelle View Post
    1. 4cos^2 3x = 1
    2. cos(2x-pi/2)=root2/2
    3. cos2x=sinx+1
    4. sin6xcosx-cos6xsinx= - root3/2
    4. cos2x-4=3cosx-4sin^2x

    Please explain to me how to solve these!!

    Also, how can i type the root or squares out like I see people do in this forum? or is it an image?
    For the mathematical typesetting, it's using an inbuilt LaTeX compiler. Click on an image and its code comes up.


    1. 4\cos^2{(3x)} = 1

    \cos^2{(3x)} = \frac{1}{4}

    \cos{(3x)} = \pm \frac{1}{2}

    3x = \arccos{\left(\pm \frac{1}{2}\right)}

    3x = \left\{\frac{\pi}{3}, \pi - \frac{\pi}{3}, \pi + \frac{\pi}{3}, 2\pi - \frac{\pi}{3}\right\} + 2\pi n, n \in \mathbf{Z}

    3x = \left \{ \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\right\} + 2\pi n

    x = \left \{ \frac{\pi}{9}, \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{5\pi}{9}\right\} + \frac{2\pi n}{3}.
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  4. #4
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    Quote Originally Posted by dorkymichelle View Post
    1. 4cos^2 3x = 1
    2. cos(2x-pi/2)=root2/2
    3. cos2x=sinx+1
    4. sin6xcosx-cos6xsinx= - root3/2
    4. cos2x-4=3cosx-4sin^2x

    Please explain to me how to solve these!!

    Also, how can i type the root or squares out like I see people do in this forum? or is it an image?
    2. \cos{\left(2x - \frac{\pi}{2}\right)} = \frac{\sqrt{2}}{2}

    2x - \frac{\pi}{2} = \arccos{\left(\frac{\sqrt{2}}{2}\right)}

    2x - \frac{\pi}{2} = \left\{ \frac{\pi}{6}, 2\pi - \frac{\pi}{6}\right\} + 2\pi n, n \in \mathbf{Z}

    2x - \frac{\pi}{2} = \left \{ \frac{\pi}{6}, \frac{11\pi}{6} \right\} + 2\pi n

    2x = \left \{ \frac{2\pi}{3}, \frac{7\pi}{3}\right\} + 2\pi n

    x = \left \{ \frac{\pi}{3}, \frac{7\pi}{6} \right\} + \pi n.
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  5. #5
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    Quote Originally Posted by dorkymichelle View Post
    1. 4cos^2 3x = 1
    2. cos(2x-pi/2)=root2/2
    3. cos2x=sinx+1
    4. sin6xcosx-cos6xsinx= - root3/2
    4. cos2x-4=3cosx-4sin^2x

    Please explain to me how to solve these!!

    Also, how can i type the root or squares out like I see people do in this forum? or is it an image?
    3. \cos{(2x)} = \sin{x} + 1

    1 - 2\sin^2{x} = \sin{x} + 1

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

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


    Therefore

    \sin{x} = 0 or 2\sin{x} + 1 = 0

    Case 1:

    \sin{x} = 0

    x = \arcsin{0}

    x = \left \{ 0, \pi \right\} + 2\pi n, n \in \mathbf{Z}.


    Case 2:

    2\sin{x} + 1 = 0

    2\sin{x} = -1

    \sin{x} = -\frac{1}{2}

    x = \arcsin{\left(-\frac{1}{2}\right)}

    x = \left \{ \pi - \frac{\pi}{6}, \pi + \frac{\pi}{6} \right \} + 2\pi n, n \in \mathbf{Z}

    x = \left \{ \frac{5\pi}{6}, \frac{7\pi}{6} \right \} + 2\pi n.


    Putting them together we get

    x = \left \{0, \frac{5\pi}{6}, \pi, \frac{7\pi}{6} \right \} + 2\pi n.
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  6. #6
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    Okay, I'll repost these with my attempts, which to warn you guys.. are horrible!
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