Results 1 to 4 of 4

Math Help - How to solve this trig equation!

  1. #1
    Junior Member
    Joined
    Nov 2010
    Posts
    58

    How to solve this trig equation!

    1. The problem statement, all variables and given/known data

    Solve for θ.
    2sinθcosθ + 1 - 2sin^2θ = 0

    2. Relevant equations

    No equations, just identities.

    3. The attempt at a solution
    I tried factoring, but I don't know how to continue with 2 diff ID's in one equation.
    I don't know how to simplify it so that everything is in terms of one similar Identity.
    :\
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    \displaystyle 2sinxcosx=sin2x

    Make the sub:

    \displaystyle sin2x+1-2sin2x=1-sin2x\rightarrow .....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2010
    Posts
    58
    Quote Originally Posted by dwsmith View Post
    \displaystyle 2sinxcosx=sin2x

    Make the sub:

    \displaystyle sin2x+1-2sin2x=1-sin2x\rightarrow .....
    I think you might have misread that I meant -2sin(square)x instead of -2sin2x, lol.

    If I subbed that
    I would get:
    -2sin^2(x) + sin2(x) + 1 = 0
    I wanted to factor, but I got stuck at the sin2(x) part. I only know how to factor if the 2 was in front...?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by TN17 View Post
    1. The problem statement, all variables and given/known data

    Solve for θ.
    2sinθcosθ + 1 - 2sin^2θ = 0

    2. Relevant equations

    No equations, just identities.

    3. The attempt at a solution
    I tried factoring, but I don't know how to continue with 2 diff ID's in one equation.
    I don't know how to simplify it so that everything is in terms of one similar Identity.
    :\
    First substitute \displaystyle 2 \sin(\theta) \cos(\theta) = \sin(2 \theta) and \displaystyle \sin^2 (\theta) = \frac{1 - \cos(2 \theta)}{2} and simplify.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Compute Trig Function Values, Solve Trig Equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: September 8th 2011, 07:00 PM
  2. Solve this trig equation
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: December 18th 2010, 06:53 AM
  3. solve trig equation help.
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 20th 2009, 06:11 AM
  4. How do I solve this trig equation??
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 16th 2009, 07:16 PM
  5. Please help me solve this trig equation
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: February 6th 2008, 07:34 PM

Search Tags


/mathhelpforum @mathhelpforum