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Trigonometric graph question

  1. Apr 13, 2007 #1
    just encountered this question and kinda confused at how to solve it since I havn't bin told and havn't worked it out for myself. hope you can help.

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

    Use the graphs (shows 2 graphs) to find the values of x in the range [tex]0 /leq x /leq 720[/tex] when 2sinx = cosx -1

    2. Relevant equations



    3. The attempt at a solution

    I found from the graph that x could equal 0, 360, and 720. but the 2 lines cross at another point before x=360 so there should be 2 more values of x to satisfy the question.

    however, i can't easily read off an accurate result. i am only 16 so do you rekon they allow for error reading off the graph instead of working it out a very accurate way?

    thnx
     
  2. jcsd
  3. Apr 13, 2007 #2

    Gib Z

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    Homework Helper

  4. Apr 13, 2007 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Graphing calculators will allow you to "zoom" in on a point so if you have one that should give you very accurate solutions. It is, however, possible to get "exact" solutions. [itex]cos x= \sqrt{1- sin^2 x}[/itex] so you can rewrite the equation as [itex]2 sin x= \sqrt{1- sin^2 x}- 1[/itex]. Now, just to simplify the writing, let y= sin x. The equation is [itex]2 y= \sqrt{1- y^2}- 1[/itex]. Add 1 to both sides: [itex] 2y+ 1= \sqrt{1- y^2}[/itex] and, finally, squaring, [itex] 4y^2+ 4y+ 1= 1- y^2[/itex] or [itex]5y^2+ 4y= y(5y+4)= 0[/itex]. One solution to that is y= sin x= 0 (and we must also have cos x= 2sin x+ 1= 1). that gives you multiples of 360 as solutions. If y is not 0 then we must have 5y+ 4= 0 so y= sin x= -4/5 (and cos x= 2sin x+ 1= -8/5+1= 3/5. x= Arcsin(-4/5)+ multiples of 360.
     
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