How can I get θ out of this equation?

[AHashemi]

Hi
I need to get θ out of this equation.
I have no Idea how. please help.

Note: mu is a constant number.

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

 

[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}.

 

[AHashemi]

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}.

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?

 

[DrClaude]

can you give me the complete answer?

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

 

[AHashemi]

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

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.

 

[BvU]

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

Homework Statement

Homework Equations

The Attempt at a Solution

​

 

[AHashemi]

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

Homework Statement

Homework Equations

The Attempt at a Solution

​

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...

 

[cnh1995]

Hi
I need to get θ out of this equation.
I have no Idea how. please help.

Note: mu is a constant number.

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

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.

Note: mu is a constant number.

Derivative of mu w.r.t.θ will be 0.

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

 

[SteamKing]

Hi
I need to get θ out of this equation.
I have no Idea how. please help.

Note: mu is a constant number.

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

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).

 

[AHashemi]

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.

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?

 

[AHashemi]

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).

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.

 

[cnh1995]

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?

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

 

[SteamKing]

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

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)

 

[cnh1995]

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)

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.

 

[cnh1995]

log (a + b) ≠ log (a) + log (b)

I know.

 

[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.

 

[AHashemi]

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.

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.

 

[NascentOxygen]

Hi
I need to get θ out of this equation.
I have no Idea how. please help.

Note: mu is a constant number.

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

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

 

[AHashemi]

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

mu is always positive and smaller than 1.

 

[NascentOxygen]

mu is always positive and smaller than 1.

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.