Results 1 to 4 of 4

Math Help - trig identity

  1. #1
    Senior Member
    Joined
    Apr 2009
    Posts
    253

    trig identity

    Solve for x in sinx tanx + tanx – 2sinx + cosx = 0 for 0≤ x ≤ 2π rad.?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,861
    Thanks
    742
    Hello, yoman360!

    Please check for typos.

    The best I can do is reduce it to a cubic equation with no rational roots.


    Solve for x\!:\;\;\sin x\tan x + \tan x - 2\sin x + \cos x \:=\: 0\;\text{ for }0 \le x \le 2\pi

    We see that neither \tfrac{\pi}{2} nor \tfrac{3\pi}{2} is a root of the equation.


    Divide by \cos x\!:\;\;\frac{\sin x\tan x}{\cos x} + \frac{\tan x}{\cos x} - \frac{2\sin x}{\cos x} + \frac{\cos x}{\cos x} \:=\:\frac{0}{\cos x}

    . . . . . . . . . . . . . . \tan^2\!x + \tan x\sec x - 2\tan x + 1 \:=\:0

    . . . . . . . . . . . . . . . . . . . . . \tan^2\!x - 2\tan x + 1 \;=\;-\tan x\sec x

    . . . . . . . . . . . . . . . . . . . . . . . . . . (\tan x - 1)^2 \;=\;-\tan x\sqrt{\tan^2\!x+1}


    Square both sides: . . . . . . . . . . . . (\tan x - 1)^4 \;=\;\tan^2\!x(\tan^2\!x + 1)

    . . . . . . . . . \tan^4\!x - 4\tan^3\!x + 6\tan^2\!x - 4\tan x + 1 \;=\;\tan^4\!x + \tan^2\!x


    And we have: . . . . 4\tan^3\!x - 5\tan^2\!x + 4\tan x - 1 \:=\;0

    . . This cubic equation has no rational roots.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I realize that the roots need not be rational.

    For example, if x = \tfrac{\pi}{6}\;(30^o), then \tan\tfrac{\pi}{6} = \tfrac{1}{\sqrt{3}} could be a root . . . (It isn't.)

    And if x = \tfrac{\pi}{12}\;(15^o), then \tan\tfrac{\pi}{12} = 2-\sqrt{3} could be a root . . . (It isn't.)

    At this point, I surrendered . . .

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Let's multiply by \cos(x). This gives :

    \sin^2(x)+\sin(x)-2\sin(x)\cos(x)+\cos^2(x)=0

    but we know that \cos^2(x)+\sin^2(x)=1 and \sin(2x)=2\sin(x)\cos(x)

    so the equation is now

    \sin(x)-\sin(2x)+1=0

    Perhaps it has a better form this way...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2009
    From
    Kingston upon Thames, Surrey
    Posts
    2
    For what it is worth, pursuing Moo's initial stages, by iteration the solution is 1.1131Pi or 3.49683 radians. There appears to be only the one real solution.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. trig identity help...
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: May 17th 2011, 10:43 AM
  2. Another trig identity
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: February 14th 2011, 05:54 PM
  3. Trig Identity
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 18th 2010, 06:43 PM
  4. Trig identity
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: March 10th 2010, 12:39 AM
  5. Odd Trig Identity
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: August 27th 2008, 09:06 PM

Search Tags


/mathhelpforum @mathhelpforum