Results 1 to 5 of 5

Math Help - Please help!

  1. #1
    Newbie
    Joined
    Mar 2006
    Posts
    17

    Please help!

    Find the general solutions of each of the following trigonometric equations:
    a) 3sin²x + cos x - 1 = 0
    b) cos 6θ + cos 2θ = 0


    Thank you for your help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    Quote Originally Posted by LilDragonfly
    Find the general solutions of each of the following trigonometric equations:
    a) 3sin²x + cos x - 1 = 0
    b) cos 6θ + cos 2θ = 0


    Thank you for your help
    a) change (sinx)^2 into 1- (cosx)^2 you will get a quadratic in cosx
    using qudratic formula you will get cosx =1, -2/3
    hence, x = [2n*pi ]and [2n*pi + cos inverse of (-2/3)]
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by LilDragonfly
    b) cos 6θ + cos 2θ = 0
    Use the formula,
    \cos x+\cos y=2\cos \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)
    Thus,
    \cos 6x+\cos 2x=0
    Use the formula,
    2\cos 4x\cos 2x=0
    Thus,
    \left\{ \begin{array}{c} \cos 4x=0\\\cos 2x=0
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2006
    Posts
    17
    Quote Originally Posted by malaygoel
    a) change (sinx)^2 into 1- (cosx)^2 you will get a quadratic in cosx
    using qudratic formula you will get cosx =1, -2/3
    hence, x = [2n*pi ]and [2n*pi + cos inverse of (-2/3)]
    Hi again, would you be able to write this in equation format so that it is easier for me to understand? Thank you
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,548
    Thanks
    541
    Hello, LilDragonfly!

    a)\;3\sin^2x + \cos x - 1 \:=\:0

    This is what Malaygoel suggested . . .


    We have: . 3\underbrace{\sin^2x} + \cos x - 1 \;= \;0

    Then: .  3\overbrace{(1 - \cos^2x)} + \cos x - 1 \;= \;0

    And we have a quadratic: . 3\cos^2x - \cos x - 2 \;= \;0

    . . which factors: . (\cos x - 1)(3\cos x + 2)\;=\;0

    . . and has two equations to solve:

    . . . . . \cos x - 1 \:=\:0\quad\Rightarrow\quad\cos x = 1\quad\Rightarrow\quad x = 0
    . . . . . 3\cos x + 2 \:=\:0\quad\Rightarrow\quad \cos x = -\frac{2}{3}\quad\Rightarrow\quad x = \cos^{-1}\left(-\frac{2}{3}\right)


    If they want all possible solutions, you can generalize . . . right?
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum