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Solving for x (Trig)

  1. Jul 9, 2008 #1
    cosx(2sinx+1) = 0 ...it looks so easy.


    method 1:

    distribute: 2sinxcosx + cosx = 0

    double angle: sin2x + cosx = 0

    sin2x = -cosx

    here I'm getting stuck.

    Method 2:

    I try to use the zero product property

    cosx = 0
    2sinx + 1 = 0

    x = pi/2 + 2pi(k) , 3pi/2 + 2pi(k)

    x = pi/6 + 2pi(k) , 5pi/6 + 2pi(k)

    Where am I going wrong?
     
  2. jcsd
  3. Jul 9, 2008 #2
    If I were to ask you to solve for the roots of x for this equation, how would you go about it?

    [tex]x(x+1)=0[/tex]

    Apply the same method and it's solved! So Method 1 should be tossed out!

    You have the correct answers for Method 2.

    [tex]\cos x=0[/tex]

    [tex]x=\frac{\pi}{2}, \frac{3\pi}{2}=\frac{\pi}{2}+k\pi[/tex]
     
  4. Jul 9, 2008 #3
    Double angle equation (trig)

    sin2x + cosx = 0

    Attempt:

    2sinxcosx + cosx = 0

    sinx = -cosx/2cosx

    sinx = -1/2

    This gives me two solutions

    x = 7pi/6 , 11pi/6 in the interval [0 , 2pi)

    But the book gives 4...
     
  5. Jul 9, 2008 #4

    rock.freak667

    User Avatar
    Homework Helper

    Re: Double angle equation (trig)


    2sinxcosx+cosx=0
    cosx(2sinx-1)=0


    Don't divide by a trig function unless they told you that cosx[itex]\neq[/itex]0

    Now you have a product. Each one is equal to zero. Solve now.
     
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