Results 1 to 10 of 10

Thread: challenge set for solving trig equations

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    106

    challenge set for solving trig equations

    challenge set for solving trig equations-arigato.png
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    a) Cross multiply, expand, move everything to one side, factorise and solve.

    b) Multiply both sides by $\displaystyle \tan{x}$ to get $\displaystyle \tan^2{x} = 1$. Solve.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, aeroflix!

    $\displaystyle (3)\;\;\cot^{\text{-}1}(2x) - \tan^{\text{-}1}(x) \:=\:30^o$

    Note that: .$\displaystyle \cot^{\text{-}1}(2x) \;=\;\tan^{\text{-}1}\!\left(\frac{1}{2x}\right)$ . . . (Think about it!)

    So we have: .$\displaystyle \tan^{\text{-}1}\!\left(\frac{1}{2x}\right) - \tan^{\text{-}1}(x) \:=\:30^o$


    Take the tangent of both sides:

    . . $\displaystyle \tan\bigg[\tan^{\text{-}1}\!\left(\frac{1}{2x}\right) - \tan^{\text{-}1}(x)\bigg] \;=\;\tan30^o$

    . . $\displaystyle \dfrac{\tan\left[\tan^{\text{-}1}(\frac{1}{2x})\right] - \tan\left[\tan^{\text{-}1}(x)\right]} {1 + \tan\left[\tan^{\text{-}1}(\frac{1}{2x})\right] \tan\left[\tan^{\text{-}1}(x)\right]} \;=\;\dfrac{1}{\sqrt{3}}$

    . . $\displaystyle \dfrac{\frac{1}{2x} - x}{1 - (\frac{1}{2x})(x)} \;=\;\dfrac{1}{\sqrt{3}} \quad\Rightarrow\quad 2\sqrt{3}\,x^2 + 3x - \sqrt{3} \:=\:0$

    Quadratic Formula: .$\displaystyle x \;=\;\dfrac{-3 \pm\sqrt{33}}{4\sqrt{3}} $


    Therefore: . $\displaystyle x \;=\;\begin{Bmatrix}\;\;0.396143496 \\ \text{-}1.262168899 \end{Bmatrix}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2009
    Posts
    106
    yes i remember it,, its the same step just like the tan 60 degrres before right???? how did u know and figure out how to solve it?? whats your trick?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Experience
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Sep 2009
    Posts
    106
    here,, i solved letter b but its wrong

    i ended up with 1/2 is not equal to 1/3

    heres my solution or scratch
    Follow Math Help Forum on Facebook and Google+

  7. #7
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by aeroflix View Post
    here,, i solved letter b but its wrong

    i ended up with 1/2 is not equal to 1/3

    heres my solution or scratch
    You're trying too hard. Recall that $\displaystyle \cot(x) = \frac{1}{\tan(x)}$

    Thus $\displaystyle \tan(x) = \frac{1}{\tan(x)}$

    Multiply by $\displaystyle \tan(x)$

    $\displaystyle \tan^2(x) = 1$ which is equal to $\displaystyle \tan^2(x)-1^2 = 0$

    Solve the quadratic, I would use the difference of two squares to solve

    Edit: Oops, I posted the answer to B not A which is in the quote




    From what I can tell you're right (although confusing lol) up to $\displaystyle \sec^2(x) =2$. However, when you took the square root you didn't take the negative.

    $\displaystyle \sec(x) = \pm \sqrt{2}$

    Taking the reciprocal of both sides: $\displaystyle \cos(x) = \pm \frac{1}{\sqrt{2}}$


    I got answers of $\displaystyle \frac{\pi}{4}$ and $\displaystyle -\frac{\pi}{4}$ (or $\displaystyle \pm 45^{\circ}$ ) as solutions (although only principal solutions because no range is defined in the question)

    $\displaystyle \sec(x) \neq 0$ so discard that solution (from factoring out $\displaystyle \sec(x)$)
    Last edited by mr fantastic; Sep 4th 2010 at 09:25 PM. Reason: Added latex tags.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    From $\displaystyle 2\sec{x} + \sec^2{x} = \sec^2{x} + \sec^3{x}$

    $\displaystyle 2\sec{x} = \sec^3{x}$

    $\displaystyle 0 = \sec^3{x} - 2\sec{x}$

    $\displaystyle 0 = \sec{x}(\sec^2{x} - 2)$

    $\displaystyle 0 = \sec{x}(\tan^2{x} - 1)$ since $\displaystyle \tan^2{x} = \sec^2{x} - 1$

    $\displaystyle \sec{x} = 0$ or $\displaystyle \tan^2{x} - 1 = 0$

    Case 1: Does $\displaystyle \sec{x}$ ever equal $\displaystyle 0$?

    Case 2: $\displaystyle \tan^2{x} - 1 = 0$

    $\displaystyle \tan^2{x} = 1$

    $\displaystyle \tan{x} = \pm 1$

    $\displaystyle x = \frac{\pi}{4} + \frac{n\pi}{2}$ where $\displaystyle n \in \mathbf{Z}$

    $\displaystyle x = \frac{\pi + 2n\pi}{4}$ where $\displaystyle n \in \mathbf{Z}$

    $\displaystyle x = \frac{(2n + 1)\pi}{4}$ where $\displaystyle n \in \mathbf{Z}$.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Sep 2009
    Posts
    106
    where did n pi / 2 came from?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Sep 2009
    Posts
    106
    what is the porper way t o solve sec^2x = sqaure root of 2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving trig equations
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Dec 8th 2010, 05:06 PM
  2. solving trig equations
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Jan 18th 2010, 06:18 AM
  3. Need help solving trig. equations.
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Oct 5th 2009, 06:13 PM
  4. Solving Trig Equations
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Sep 5th 2009, 06:34 PM
  5. Replies: 1
    Last Post: Sep 29th 2008, 06:46 AM

Search Tags


/mathhelpforum @mathhelpforum