Trig equation: $a \sin \frac{a \pi}{2} = 1$ How do I solve the following? I am having a bit of a slow moment. 
$$a \sin \frac{a \pi}{2} = 1$$
 A: Your equation has the form 
$$\sin \left(\frac{\pi x}{2}\right)=\frac{1}{x}$$. You cannot find all solutions with a closed form, but you can see the solutions in a graph if you represents the two functions 
$$
y=\sin \left(\frac{\pi x}{2}\right) \quad and \quad y=\frac{1}{x}
$$
These two graphs are simple to plot and the abscissas of the intersection points are the solutions of the equation. From the graph it is simple to see the two solutions $x=\pm1$ and to estimate the intervals in wich the other solutions are contained. 
A: This equation has infinitely many solutions. We can approximately recover some of the solutions, but there is no general way of writing them all down.
First of all, we know that $\sin{\frac{\pi a}{2}}$ is continuous and periodic with period $T=4$, so it takes on each value in $(-1,1)$ twice in each period. In particular it will take on the value $\frac{1}{a}$ at least twice in each interval $a \in [x,x+4]$.
We can write down an asymptotic series for the roots of this equation when $\lvert a \rvert$ is large by noticing that as $\lvert a \rvert$ grows larger and larger, $\sin{\frac{\pi a}{2}}$ must get closer and closer to $0$ for the product to be equal to $1$. Thus, the solution $a_m$ becomes closer and closer to an even integer, where $\sin{\frac{\pi a}{2}}$ is $0$.
So, I postulate that the roots of this equation have an asymptotic form $$a_m = 2 m + \sum_{n=1}^{\infty} c_n \bigg(\frac{1}{m}\bigg)^n,$$ with the $c_n$ some coefficients, then we can expand $\sin{\frac{\pi a}{2}}$ as a Tayor series near the point $a = 2m$, $$a \sin{\frac{\pi a}{2}} \sim_{a \to 2m} m \pi \cos({m \pi}) (a_m - 2m) + \mathcal{O}(a_m - 2m)^2 = 1.$$ If we neglect the higher order terms we can solve this linear equation to find $$a_m = 2m + \frac{(-1)^m}{m \pi}.$$
Next, we repeat this procedure, expanding $a \sin{\frac{\pi a}{2}}$ as a Taylor series around the new estimate of the roots $a_m = 2m + \frac{(-1)^m}{m \pi}.$ We keep only the first term in the Taylor expansion and solve the resulting linear equation to find 
$$ a_m = 2m + \frac{(-1)^m}{m \pi} + \frac{(-1)^m \pi - 12}{24 \pi^2 m^3} + \mathcal{O}\bigg(\frac{1}{m}^4\bigg). $$
One can repeat this process indefinitely because it only involves repeatedly solving a linear equation for $a_m$, but then the computations get progressively messier as you go to higher and higher orders. 
This computation is surprisingly accurate, for instance even for the 3rd positive root of your equation (which corresponds to m=2), the difference between the exact answer and the 3rd order approximate answer is $$x_{3} - x_{m=2} \approx  -0.00027.$$
Note: Also be aware that one could calculate the first $t$ terms in such an expansion and then use Richardson extrapolation, or some other convergence acceleration technique, to get even more precise approximate answers without generating any additional terms in the expansion.
