# How can I get θ out of this equation?

1. Dec 4, 2015

### AHashemi

Hi
I need to get θ out of this equation.

Note: mu is a constant number.

$-6\mu sin\theta - 3cos\theta + 2(((e^2)^\theta)^\mu)*(1-2\mu^2)=0$

2. Dec 4, 2015

### HallsofIvy

Start by writing sine and cosine in terms of the exponential: $sin(\theta)= \frac{e^{i\theta}- e^{-i\theta}}{2i}$ and $cos(\theta)= \frac{e^{i\theta}+ e^{-i\theta}}{2}$.

3. Dec 4, 2015

### AHashemi

thanks but this just made things harder for me. now I have to deal with i too. this equation is the result of a classical mechanic problem where I need to find theta.
and mu is a positive number.
can you give me the complete answer?

4. Dec 4, 2015

### Staff: Mentor

Is this homework? Even if it isn't, you can't expect other people to do your work for you.

5. Dec 4, 2015

### AHashemi

No it's not. I'm trying to solve a problem without removing friction.
But here's the case: I'm studying elementary physics (classical mechanics) currently but the math part of this problem is way advanced than my knowledge so I need help.

6. Dec 4, 2015

### BvU

Perhaps it's a good idea to complete the template below ?
I'm curious how things can have become so complicated !

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

7. Dec 4, 2015

### AHashemi

1,2. I've already given the equation.

3. this equation was half a page long and this was the shortest form of it I could get to. and I have no Idea how to solve it.

I think it's completely right to ask for help where I'm sure I can't do anything more.
I'm also curious how can you make things this complicated...

8. Dec 4, 2015

### cnh1995

You can rewrite the equation by taking all the trigonometric terms on one side. Then you can separate the terms by taking natural logarithm on both sides.
Derivative of mu w.r.t.θ will be 0.

I don't know if that'll work..its just a suggestion.

9. Dec 4, 2015

### SteamKing

Staff Emeritus
If by getting θ out of the equation, you mean finding the value of θ which satisfies the equation, I think the quickest way is for various values of μ, you iterate the equation with different values of θ until you find a value which makes the left hand side equal to the right hand side (i.e., zero).

You can set this calculation up on a spreadsheet. For faster convergence, you might employ Newton's method, if you can take the derivative of f(θ).

10. Dec 4, 2015

### AHashemi

the problem is mu and theta are positive numbers and if I move e to on side of the equation It would be negative and we can't apply natural logarithm. right?

11. Dec 4, 2015

### AHashemi

By getting theta out I mean I want theta on one side of the equation and the rest in the other side. like this:
θ = something (not theta itself)

so I can give value of mu and find theta whenever.

12. Dec 4, 2015

### cnh1995

You keep e as it is and move the trigonometric part on the RHS. That way you can take natural log on both sides.

13. Dec 4, 2015

### SteamKing

Staff Emeritus
Yeah, but you still have the log of the sum of a couple of trig functions on one side.

Remember, log (a + b) ≠ log (a) + log (b)

14. Dec 4, 2015

### cnh1995

That's why I suggested to take derivative of the terms w.r.t.θ to eliminate the constants but I now think it won't work. Its not the right approach.

15. Dec 4, 2015

### cnh1995

I know.

16. Dec 4, 2015

### FactChecker

If this is not a homework problem, why do you think there is a closed form solution for θ? Real-world problems are often not that simple. You may need to use numerical techniques like Newton's method to estimate the solution.

17. Dec 5, 2015

### AHashemi

I showed the equation to a physics professor and he said nearly the same thing that you can't always have the requested parameter in one side of the equation.

Thanks again everyone.

18. Dec 6, 2015

### Staff: Mentor

What is the expected range of μ for which you need to apply this?

19. Dec 6, 2015

### AHashemi

mu is always positive and smaller than 1.

20. Dec 8, 2015

### Staff: Mentor

I was asking whether you required the equation over a limited range of μ, because were this so then it's often possible to replace a complicated expression with a much simpler one which, for some limited range, gives results still as accurate as you'd wish. Example, for μ in the range, say, 0.6 to 0.8, you might find that a portion of a parabola would be a good approximation for your purposes.

But where you require the full range of μ, then staying with the original equation is best.