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.
 A: One degree is $\frac{\pi}{180}$ radians, so what we want here is
$$\sin(x)=\sin(\frac{\pi x}{180})$$
And so
$$x = \frac{\pi x}{180}+2\pi k, \qquad \text{or} \qquad x = \frac{-\pi x}{180}+\pi (2k+1)$$
Where $k$ is any integer (since $\sin(a) = \sin(b)$ iff $a-b$ is an integer multiple of $2\pi$, or if $a+b$ is an odd multiple of $\pi$). Solving for $x$,
$$x(1 - \frac{\pi}{180}) = 2\pi k, \qquad \text{or} \qquad x(1 + \frac{\pi}{180}) = \pi (2k+1)$$
And now we just need to find the $k$ that gives us the smallest positive solution. Since the number on the left hand side is positive (in both cases), we choose $k = 1$ for the left, and $k = 0$ for the right. Then
$$x = \frac{2\pi}{1-\frac{\pi}{180}}, \qquad \text{or} \qquad x = \frac{\pi}{1+\frac{\pi}{180}}$$
The right hand solution is smaller, so that's our answer. 
A: 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,
$$\min \left( \frac{2\pi}{1-\beta}, \frac{\pi}{1+\beta} \right) = \min \left( \frac{2\pi}{1-\frac{\pi}{180}}, \frac{\pi}{1+\frac{\pi}{180}} \right) = \frac{\pi}{1+\frac{\pi}{180}} = \left(\frac{1}{\pi}+\frac{1}{180}\right)^{-1}$$
