Answer
$tan~\alpha = \alpha$
Work Step by Step
$I(\theta) = I_m~(\frac{sin~\alpha}{\alpha})^2$
We can differentiate this expression with respect to $\alpha$:
$\frac{dI}{d\alpha} = 2I_m~(\frac{sin~\alpha}{\alpha})(\frac{\alpha~cos~\alpha-sin~\alpha}{\alpha^2}) = 0$
One possible set of solutions is:
$\frac{\alpha~cos~\alpha-sin~\alpha}{\alpha^2} = 0$
$\alpha~cos~\alpha-sin~\alpha = 0$
$\alpha~cos~\alpha = sin~\alpha$
$\frac{sin~\alpha}{cos~\alpha} = \alpha$
$tan~\alpha = \alpha$
