Smallest number whose $\sin(x)$ in radian and degrees is equal Question: 

What is the smallest positive real number $x$ with the property that the sine of $x$ degrees is equal to the sine of $x$ radians?

My try: 0. But zero isn't a positive number. How do I even begin to solve it? I tried taking inverse on both sides of $\sin \theta = \sin x$, but that didn't help.

We have
$$\sin (x) = \sin (\beta x)$$
where $\beta := \frac{\pi}{180}$. Using
$$\sin (\alpha x) = \frac{e^{i \alpha x} - e^{-i \alpha x}}{2i}$$
we conclude that the equation $\sin (x) = \sin (\beta x)$ can be rewritten as follows
$$2 \, \sin \left( \left(\frac{1-\beta}{2}\right) x\right) \, \cos \left( \left(\frac{1+\beta}{2}\right) x\right) = 0$$
The smallest positive solution is, thus,
$$\min \left( \frac{2\pi}{1-\beta}, \frac{\pi}{1+\beta} \right) = \min \left( \frac{2\pi}{1-\frac{\pi}{180}}, \frac{\pi}{1+\frac{\pi}{180}} \right) = \frac{\pi}{1+\frac{\pi}{180}} = \left(\frac{1}{\pi}+\frac{1}{180}\right)^{-1}$$
