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Homework Help: How do I solve for x in trig, functions?

  1. Jul 14, 2014 #1
    1. The problem statement, all variables and given/known data
    4/PI = 2sin(x) - sin(2x)

    2. Relevant equations

    4/PI = 2sin(x) - sin(2x)

    3. The attempt at a solution

    I have no clue how to do this. This is part of an integration problem for average values. I need to solve for f(c). I know the answer is 1.238, and 2.808.
  2. jcsd
  3. Jul 14, 2014 #2
    I don't think you can find solutions to that equation using elementary tools and the main difficulty is that you can't transform that expression to one which contain only one trigonometric function or one of the form trigonometric function expression times another trigonometric function expression equals 0, even do a closed form exists which was obtained through lots of hardly guessable substitutions, it is still too ugly and complicated. So the best thing you can use are numerical approximations which can be found in the link I referenced to.
    Last edited: Jul 14, 2014
  4. Jul 14, 2014 #3


    Staff: Mentor

    You could simplify it a bit via sin(2x) is 2sin(x)cos(x0 but its still not solvable via simple algebra. A numerical solution would probably be the best like try newton's approximation to it...
  5. Jul 15, 2014 #4


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    Science Advisor

    You can, at least, reduce it to solving a fourth degree polynomial equation. We can, as said above, write sin(2x)= 2 sin(x)cos(x) making the equation [itex]2sin(x)- 2sin(x)cos(x)= 4/\pi[/itex]. Now, replace cos(x) with [itex]\sqrt{1- sin^2(x)}[/itex]: [itex]2sin(x)- 2sin(x)\sqrt{1- sin^2(x)}= 4/\pi[/itex]. [itex]2sin(x)\sqrt{1- sin^2(x)}= 2sin(x)- 4/\pi[/itex]. [itex]sin(x)\sqrt{1- sin^2(x)}= sin(x)- 2/\pi[/itex].

    Now get rid of the square root by squaring both sides: [itex]sin^2(x)(1- sin^2(x))= sin^2(x)- 4sin(x)/\pi+ 4/\pi^2[/itex]. [itex]sin^4(x)+ 4 sin(x)/\pi- 16/\pi^2= 0[/itex].

    Let y= sin(x) and that becomes [itex]y^4+ (4/\pi)y- (4/\pi)^2= 0[/itex].
    Last edited by a moderator: Jul 15, 2014
  6. Jul 15, 2014 #5
    Thanks everyone for the input.
  7. Jul 15, 2014 #6

    D H

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    Staff Emeritus
    Science Advisor

    There are three things wrong here:
    1. This is close to a complete solution. Please don't do that, Halls.
    2. This is an incorrect complete solution. There is a math error.
    3. Even after fixing the math error, this approach admits spurious solutions.

    The basic idea of reducing the problem to a fourth degree polynomial is a good one. A way around the spurious solutions problem is to use both double and half angle formulae. Rather than running this to ground, I'll leave a couple of hints:
    • Use the identity [itex]1-\cos x = 1 - \cos\left(2 \frac x 2\right) = 2 \sin^2\frac x 2[/itex].
    • Use the identity [itex]\sec^4 \theta = (\sec^2\theta)^2 = (1+\tan^2\theta)^2[/itex].
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