How do I solve for x in trig, functions?

[shreddinglicks]

Homework Statement

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

Homework Equations

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

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.

 

[HakimPhilo]

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.

 

[jedishrfu]

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...

 

[HallsofIvy]

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 $2sin(x)- 2sin(x)cos(x)= 4/\pi$. Now, replace cos(x) with $\sqrt{1- sin^2(x)}$: $2sin(x)- 2sin(x)\sqrt{1- sin^2(x)}= 4/\pi$. $2sin(x)\sqrt{1- sin^2(x)}= 2sin(x)- 4/\pi$. $sin(x)\sqrt{1- sin^2(x)}= sin(x)- 2/\pi$.

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

Let y= sin(x) and that becomes $y^4+ (4/\pi)y- (4/\pi)^2= 0$.

 

[shreddinglicks]

Thanks everyone for the input.

 

[D H]

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 $2sin(x)- 2sin(x)cos(x)= 4/\pi$. Now, replace cos(x) with $\sqrt{1- sin^2(x)}$: $2sin(x)- 2sin(x)\sqrt{1- sin^2(x)}= 4/\pi$. $2sin(x)\sqrt{1- sin^2(x)}= 2sin(x)- 4/\pi$. $sin(x)\sqrt{1- sin^2(x)}= sin(x)- 2/\pi$.

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

Let y= sin(x) and that becomes $y^4+ (4/\pi)y- (4/\pi)^2= 0$.

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 $1-\cos x = 1 - \cos\left(2 \frac x 2\right) = 2 \sin^2\frac x 2$.
• Use the identity $\sec^4 \theta = (\sec^2\theta)^2 = (1+\tan^2\theta)^2$.