Really complicated solve for x problem

[phys-lexic]

really complicated "solve for x" problem.. please help..

Homework Statement

[This is the final step in a "critical thinking" problem assigned as extra practice/intense application] Find the value of x, for the given equation, when f(x) = \frac{49}{6}\pi


f(x) = \left(x\right)\times\sqrt{49-x^2} + 49sin^{-1}\left(\frac{x}{7}\right)

Homework Equations

(This is where I need help, I have tried moving around the values, sqaring both sides, applying e and ln; my T.A. could only think of plugging f(x) into a graphing calculator and tracing to y = \frac{49}{6}\pi)
*A big question I have is if trig-substitution (aside from integration) can be used, or another method I am not "equipped with," with simplifications.

The Attempt at a Solution

This is what is left after integrating a problem, the answer should be ~1.85 (from graphing/tracing). I tried simplifying using regular relationships:

sin^{-1}\left(\frac{x}{7}\right) = \frac{1}{6}\pi - \left(x\sqrt{49-x^2}\right)\div49

 

[Mark44]

You're not going to be able to solve this by algebraic means. The simplest approach is to graph the function and see what value of x gives a y value of 49pi/6.

 

[tiny-tim]

Hi phys-lexic!

Try the obvious substitution. ​

 

[phys-lexic]

I understand algebraic means won't help, which is why I'm posting this question.

Trig-substitution is what I was thinking, but is that applicable when not integrating? (We were only introduced to trig-substitutions with integrals, for obvious reasons)

 

[tiny-tim]

Trig-substitution is what I was thinking, but is that applicable when not integrating?

Yes! You can always substitute, if you think it will make the problem easier.

 

[diggy]

Would be a lot simpler if there was only a way to make that first term go to zero...

 

[vin300]

If f(x)=y, substitute Sqrt[a^2-x^2]=dy/dx, then it reduces to the standard form dy/dx +Py=Q

 

[phys-lexic]

Aah.. I tried it out a few times and ended up going in circles. Thankyou anyways everyone, now I know why this question is "way harder than the exam would be."