How to solve $20t + \frac{240}{\pi} \sin \frac{\pi t}{12} =2500$ by hand? I'm trying to solve a calc 1 problem about how long it takes to fill a reservoir.  This is the result of my integration.
$$20t + \frac{240}{\pi} \sin\frac{\pi t}{12}=2500$$ 
I know the answer is $t = 122.6$ (its an odd book problem), I only figured it out because I used my graphing calculator and found the intersection of the function and $y=2500$.
How can I solve for $t$ by hand?
 A: Scale down the equation as
$$t + \frac{12}{\pi} \sin \left( \frac{\pi t}{12} \right)=125$$
Note that the RHS is much larger than the sine term. So, the solution is roughly $t=125$, which can be refined by adding the first-order approximation as follows.
Let $f(t) = t -125 + \frac{12}{\pi} \sin \frac{\pi t}{12} $ and the root $t=125 + \Delta t$. Then,
$$0 = f(125) + f’(125)  \Delta t$$ 
Solve for $ \Delta t$
$$ \Delta t = - \frac{f(125)}{f’(125)} 
= -\frac{12}{\pi} \frac{\sin \frac{5\pi}{12}}{1+\cos \frac{5\pi}{12}  }
= -\frac{12}{\pi} \tan \frac{5\pi}{24}$$
Thus, the approximate analytic solution is 
$$t= 125 -\frac{12}{\pi} \tan \frac{5\pi}{24}\approx 122.1$$
A: If by hand you mean without a calculator at all, you need to use some numeric approach.  Because the sine is limited to $\pm 1$ that term cannot be $80$ in absolute value, so we can get a first approximation by ignoring it, which gives $t=125$.  You will need some way to evaluate the sine, but presumably you have a set of tables.  Now I would use fixed point iteration, writing $$t=125-\frac {12}\pi \sin\left(\frac {\pi t}{12} \right)$$
Plug in $125$ on the right, calculate the new $t$ and iterate.  It takes a while to converge because the derivative of the right side is close to $-1$.  Newton's method will converge more quickly at the price of being a bit harder to derive.
A: Consider that you look for the zero of function $$f(t)=t + \frac{12}{\pi} \sin \left( \frac{\pi t}{12} \right)-125$$ and notice that $f(120)=-5$.
So use Taylor expansion
$$f(t)=-5+2 (t-120)+O\left((t-120)^3\right)$$ Ignoring the higher order terms, an approximation is then $t=122.5$. This is the same as the first iteration of Newton method.
Sooner or later, you will learn that, for approximating functions, Padé approximants are much better than Taylor series. Using the $[1,2]$ Padé approximant (this is the simplest), the solution will then be
$$t=120+\frac{17280}{6912-25 \pi ^2}$$ Now, remembering that $\pi^2 \sim 10$, this reduces to
$$t=120+\frac{8640}{3331}=120+\frac{8640\times 3}{3331\times 3}=120+\frac{25920}{9993}\sim 120+\frac{25920}{10000}=122.592$$ while the exact solution is $122.598$.
