If $0^\circ\leqslant x<360^\circ$, what is the maximum number of solutions to the equation $\sin x = a$ where a is a real number? I tried solving the question, but I kept getting $5$ solutions. My book only has $4$ choices: $0$, $1$, $2$, or $3$ solutions. My solutions were $0^\circ$, $90^\circ$, $150^\circ$, $180^\circ$, and $270^\circ$. What did I do wrong? Why is $2$ solutions the correct answer? 

For intuition, @Mark Bennet's visual approach in the comments is very helpful.
The solution of $\sin(x) = a$ can have a different number of solutions, depending on the value of $a$:

*

*$-1<a<1$ : 2 solutions

*$a=1$ or $a=-1$: 1 solution (either $x = 90^\circ$ or $270^\circ$, since we only get the local maximum or the minimum)

*$a>1$ or $a<-1$: 0 solutions, since $\sin$ only takes values in $[-1,1]$ for real arguments.

Answer: Hence, the maximum number of solutions is $2$.
Visualization for $a = -0.3$ (green), $a = 0.7$ (orange), $a = 1$ (red), and $a = 1.3$ (purple). 
The $x$-value(s) of the intersection(s) of a horizontal line at height $a$ with the sine-graph are the desired solutions of $\sin x = a$.
The number of solutions for given $a$ is the number of times the corresponding line intersects the sine-graph.
