Results 1 to 3 of 3

Math Help - General sloution but requires manipulation

  1. #1
    Member
    Joined
    Jan 2009
    From
    London
    Posts
    92

    General sloution but requires manipulation

    I'm have troubles with finding the general solution for [tex] \theta [\math] of:
     \cos(6\theta) + \cos(3\theta) + 4\cos^2 (\frac{3}{2}\theta)sin^2(\frac{3}{2}\theta) = 0


    I know the cosine general formula is this:
    \pm\theta +2n\pi
    and that you can use identities to break the functions down until each factor can be zeroed and hence find these values. But the two additions are throwing me at the start amongst other things.

    Your help would be greatly appreciated.

    D
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    You could start by using the double angle identity for sine.

    \sin2A = 2\sin A\cos A

    Squaring both sides give:

    \sin^22A = 4\sin^2A\cos^2A

    Subtituting A by 3theta/2;

    \sin^22(\dfrac{3\theta}{2}) = 4\sin^2(\dfrac{3\theta}{2})\cos^2(\dfrac{3\theta}{  2})

    \sin^23\theta = 4\sin^2(\dfrac{3\theta}{2})\cos^2(\dfrac{3\theta}{  2})

    Now we see that we have cos 3theta and cos 6theta. Break the cos 6theta into half angle.

    \cos2A = 2\cos^2A - 1

    \cos6\theta = 2\cos^23\theta - 1

    Can you finish this now?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,803
    Thanks
    692
    Hello, dojo!

     \cos 6\theta + \cos3\theta + 4\cos^2(\tfrac{3}{2}\theta)\sin^2(\tfrac{3}{2}\the  ta) \:=\: 0

    \text{We have: }\;\cos6\theta + \cos3\theta + \underbrace{\bigg[2\sin(\tfrac{3}{2}\theta)\cos(\tfrac{3}{2}\theta)\  bigg]^2} \;=\;0
    . . . . . \overbrace{(2\cos^2\!3\theta - 1)} \;+\; \cos3\theta \;+\; \sin^23\theta \;=\;0

    . . . . . 2\cos^23\theta - 1 + \cos3\theta + \overbrace{(1 - \cos^23\theta)} \;=\;0

    . . . . . . . . . . . . . . . . . . . . \cos^23\theta + \cos3\theta \;=\;0

    . . . . . . . . . . . . . . . . . . \cos3\theta(\cos3\theta +1) \;=\;0


    Therefore: . \begin{Bmatrix}\cos3\theta \:=\: 0 & \Rightarrow & 3\theta \:=\: \frac{\pi}{2} + \pi n & \Rightarrow & \theta \:=\: \frac{\pi}{6} + \frac{\pi}{3}n \\ \\[-3mm]<br />
\cos3\theta\:=\:\text{-}1 & \Rightarrow & 3\theta \:=\:\pi + 2\pi n & \Rightarrow & \theta \:=\:\frac{\pi}{3} + \frac{2\pi}{3}n<br /> <br />
\end{Bmatrix}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: September 23rd 2011, 08:34 AM
  2. [SOLVED] Identity and general sloution problem
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: May 22nd 2010, 07:52 AM
  3. sloution?
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: August 8th 2009, 10:47 AM
  4. Evaluate sum(requires use of complex numbers)?
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: May 1st 2009, 12:11 AM
  5. Replies: 0
    Last Post: April 28th 2009, 06:09 AM

/mathhelpforum @mathhelpforum