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Given that f(x)=sin(2x), find x such that f(2x+1)=1

  1. Feb 3, 2007 #1

    2. Relevant equations

    3. The attempt at a solution

    That is how I am trying to solve to this equation, by using the compound angle formula and the double angle formula, however it seems as i am going around in circles. The correct answer for this problem is equal to:
    (pi-4)/8 + npi/2
    Is there a simpler way of solving this equations and ones which are simpler, or am trying to solve it using the correct method. All help is greatly appreciated and many thanks in advance,

  2. jcsd
  3. Feb 3, 2007 #2

    D H

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    What are the solutions to [itex]\sin \theta = 0[/itex]?
  4. Feb 3, 2007 #3
    the solution to sin(theta)=0 is sin^-1 0=0
    I am not sure how this may help, sorry :(

    I know that the graph of sin(x) oscillates between -1 and 1 and that the maximums of the graph occur at (pi/2)+n2pi, where n is a whole number variable.
  5. Feb 3, 2007 #4
    How about this as a hint:

    what is arcsin(1)?
  6. Feb 3, 2007 #5
    the arcsin(1)=pi/2
  7. Feb 3, 2007 #6
    now what is the arcsin(sin(4x+2))?
  8. Feb 3, 2007 #7
    um......4x+2 ?
  9. Feb 3, 2007 #8
    see where I am heading with this...

    Note that if you took the arcsin of the probem given you get this:


    Now just solve for your varible.

    (wasn't that much easier than trying to expand the expression)
    Last edited: Feb 3, 2007
  10. Feb 3, 2007 #9
    What about all the other solutions? How about:

    sin(4x+2) = 1= sin(pi*(2*n+1/2)), n=...-3,-2,-1,0,1,2,3....


    4x+2 = pi*(2*n+1/2), n=...-3,-2,-1,0,1,2,3....

    Now solve for x in terms of n.
  11. Feb 3, 2007 #10

    Yeah that method of looking at it works also.

    However, you can always just note that because it is an oscillating function with a period of 2pi, that you just add the 2npi statement to the end of your answer to make it general.
  12. Feb 3, 2007 #11
    Oh I practically understand the problem now. By taking the arcsine of sin(4x+2) I am left with what is in the brackets. But how come what is in the brackets (4x+2) is equal to pi/2? Does this method of taking the inverse of the trig function work for the other trig functions as well? Thank you very much for the help so far :)
  13. Feb 3, 2007 #12


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    Look, you know that:

    [tex] \sin(4x + 2) = 1 [/tex]

    and you also know for what values of [itex] \theta [/itex] a sine function is equal to 1, in general:

    [tex] \sin(\theta) = 1, \ \ \ \theta = (n+\frac{1}{2})\pi, \ \ n = 0, 1, 2, ... [/tex]

    in other words, the *argument* of the sine function must be an odd integer multiple of pi/2 if the sine is to be equal to one. Now you just equate the arguments so that satisfy that condition.
  14. Feb 4, 2007 #13
    If you take the the arcsin (or anyother inverse trig function, depending on the situation), you have to do it to both sides of the equation. Thats why it is equated to pi/2 in this situation.

    And yes, this methods works for the other trig functions (Say for instance you had cos(2x-1)=1.4, you could use the same method we have used so far to solve this problem, with a little adaptation).

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