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: Soloving sqrt(5)Cos(2x + 0.464)

  1. Jun 5, 2008 #1
    1. The problem statement, all variables and given/known data

    Solve for the interval 0 < x < pi/2


    2. Relevant equations

    3. The attempt at a solution

    Am i right in saying as cos oscillate between -1 and 1:

    The maximum is sqrt(5) and occurs when (2x-0.464) = 1

    Now i got 0.464/2 and (2pi - 0.463)/2
    and the back of the book says 0.32 and 2.36

    How do you solve equations like above :S


    yes my calculator is in radians.
  2. jcsd
  3. Jun 5, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    FIrst of all, what is it that you are trying to solve?
  4. Jun 5, 2008 #3
    oops i forgot.

    solve for x in the range stated above

    Thanks ;)
  5. Jun 5, 2008 #4


    User Avatar
    Science Advisor

    In order to solve any equation, you first have to have an equation. What equation are you talking about. Exactly what is the question you want to answer?

    Yes, the maximum of the expression you give is sqrt(5) but it does not occur when 2x+ 0.464= 1, it occurs when 2x+0.464= 0 or a multiplie of 2pi which is, I presume, where you got the values of 0.464/2 and (2pi- 0.463)/2 (it should be -0.464/2= -0.232 and (2pi 0.463)/2). But what is the question to which your book says the answer is 0.32 and 2.36?
  6. Jun 5, 2008 #5
    oops i forgot.

    solve for x in the range stated above

    Thanks ;)
  7. Jun 5, 2008 #6
    The rest of the question though is:

    I) Express 2Cos2x-Sin2x in the form RCos(2x+a) giving R exactly and a to 3dp
    II) Hence find the x-coordinates of the points of interestction of C1 and C2 in the interval 0<=x=<pi. Give your answer as radians and to 2dp

    Note that c1 and c2 are curves with the equations y = cos2x - 2sinĀ²x and y = sin2x respectively.
  8. Jun 5, 2008 #7
    it' ok i got it. cos-1(1/sqrt(5))

    easy peasy really :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook