Results 1 to 9 of 9

Math Help - Pre Calc: Quad. Trig. Equations

  1. #1
    Newbie
    Joined
    Nov 2007
    Posts
    18

    Pre Calc: Quad. Trig. Equations

    So I went pretty smoothly through most of the homework questions untill I reached these ones. I'd love some quick help since I have a test on this tomorrow!

    1) Find solutions on the interval [0, 2pi]:

    4sinxtanx - 3tanx + 20sinx - 15 = 0

    2) Use an identity to find solutions on the interval [0, 2pi]:

    sec^2x - 2tan^2x = 0

    Any hints/clues/help/ANSWERS would be greatly appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by mayflower29 View Post
    2) Use an identity to find solutions on the interval [0, 2pi]:

    sec^2x - 2tan^2x = 0

    Any hints/clues/help/ANSWERS would be greatly appreciated.
    remember \sec^2 x = 1 + \tan^2 x
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,862
    Thanks
    743
    Hello, mayflower29!

    1) Find solutions on the interval [0,\,2\pi]
    . . 4\sin x\tan x - 3\tan x + 20\sin x - 15 \:= \:0

    Factor: . \tan x(4\sin x - 3) + 5(4\sin x - 3) \;=\;0

    Factor: . (4\sin x-3)(\tan x + 5) \:=\:0

    Then: . 4\sin x-3 \:=\:0 \quad\Rightarrow\quad \sin x \:=\: \frac{3}{4} \quad\Rightarrow\quad x \:=\: \sin^{-1}\left(\frac{3}{4}\right) \;=\;\begin{Bmatrix}0.8481 \\ 2.2935\end{Bmatrix}

    . . . . . \tan x+5 \:=\: 0 \quad\Rightarrow\quad \tan x \:=\:\text{-}5 \quad\Rightarrow\quad x \:=\:\tan^{-1}(\text{-}5)\;=\;\begin{Bmatrix}1.7682 \\ 4.9100\end{Bmatrix}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2007
    Posts
    18
    Thanks a lot Soroban! That was perfect!

    Quote Originally Posted by Jhevon View Post
    remember \sec^2 x = 1 + \tan^2 x
    Is it? I don't understand where that came from. We usually use our list of identities, and the only thing to do with secx is that its 1/cosx.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by mayflower29 View Post
    Is it? I don't understand where that came from. We usually use our list of identities, and the only thing to do with secx is that its 1/cosx.
    this is another identity, and it is the appropriate one to use here
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    Quote Originally Posted by Jhevon View Post
    this is another identity, and it is the appropriate one to use here
    Or you could convert the equation to sines and cosines and work from there.

    sec^2(x) - 2tan^2(x) = 0

    \frac{1}{cos^2(x)} - \frac{2sin^2(x)}{cos^2(x)} = 0

    \frac{1 - 2sin^2(x)}{cos^2(x)} = 0

    Now multiply both sides by cos^2(x):
    1 - 2sin^2(x) = 0

    etc.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Nov 2007
    Posts
    18
    Quote Originally Posted by topsquark View Post
    Or you could convert the equation to sines and cosines and work from there.

    sec^2(x) - 2tan^2(x) = 0

    \frac{1}{cos^2(x)} - \frac{2sin^2(x)}{cos^2(x)} = 0

    \frac{1 - 2sin^2(x)}{cos^2(x)} = 0

    Now multiply both sides by cos^2(x):
    1 - 2sin^2(x) = 0

    etc.

    -Dan
    Makes sense, thanks a lot!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    aziz0917
    Guest
    Quote Originally Posted by Soroban View Post
    Hello, mayflower29!


    Factor: . \tan x(4\sin x - 3) + 5(4\sin x - 3) \;=\;0

    Factor: . (4\sin x-3)(\tan x + 5) \:=\:0

    Then: . 4\sin x-3 \:=\:0 \quad\Rightarrow\quad \sin x \:=\: \frac{3}{4} \quad\Rightarrow\quad x \:=\: \sin^{-1}\left(\frac{3}{4}\right) \;=\;\begin{Bmatrix}0.8481 \\ 2.2935\end{Bmatrix}

    . . . . . \tan x+5 \:=\: 0 \quad\Rightarrow\quad \tan x \:=\:\text{-}5 \quad\Rightarrow\quad x \:=\:\tan^{-1}(\text{-}5)\;=\;\begin{Bmatrix}1.7682 \\ 4.9100\end{Bmatrix}


    Sorry, i was just wondering how did you get the second factor part you did.The (tanx+5 part) Can you please explain to me how you did that? Thankss
    Follow Math Help Forum on Facebook and Google+

  9. #9
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by aziz0917 View Post
    Sorry, i was just wondering how did you get the second factor part you did.The (tanx+5 part) Can you please explain to me how you did that? Thankss
    note that in the first line of what you quoted, the 4 \sin x - 3 was a common term, so he factored it out. the result was that a factor of \tan x + 5 emerged
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. quad. equations
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 25th 2011, 08:47 AM
  2. pre-calc/trig
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: December 18th 2008, 12:10 AM
  3. Pre-Calculus - Quad/Poly Equations and Circles
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: September 7th 2008, 11:50 PM
  4. trig with no calc
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: August 27th 2008, 03:57 PM
  5. What am I doing? Calc, Pre-calc, Trig, etc...
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: April 23rd 2008, 09:51 AM

Search Tags


/mathhelpforum @mathhelpforum