1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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?

  2. jcsd
  3. Apr 13, 2007 #2

    Gib Z

    User Avatar
    Homework Helper

  4. Apr 13, 2007 #3


    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook