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? 
 A: Clearly, we need $-1\le a\le1$ for at least one real solution
If $\sin x_1=\sin x_2$
Using Prosthaphaeresis Formulas, $$\sin x_1-\sin x_2=2\sin\dfrac{x_1-x_2}2\cos\dfrac{x_1+x_2}2.$$
If $\sin\dfrac{x_1-x_2}2=0\implies\dfrac{x_1-x_2}2=m180^\circ\iff x_1\equiv x_2\pmod{360^\circ}.$
If $\cos\dfrac{x_1+x_2}2=0\implies\dfrac{x_1+x_2}2=(2n+1)90^\circ\iff x_1\equiv 180^\circ- x_2\pmod{360^\circ}$
Now $x_1,x_2$ will coincide if $x_1\equiv 180^\circ- x_1\pmod{360^\circ}\iff x_1\equiv90^\circ\pmod{180^\circ}.$
In that case, we shall have only one in-congruent solution $\pmod{360^\circ}.$
Otherwise, there will be distinct two namely, $x_1,180^\circ- x_1.$
A: 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.
A: By definition, the sine of an angle $\theta$ is the $y$-coordinate of a point where the line through the origin that makes an angle of $\theta$ to the $x$-axis intersects the unit circle.  That's kind of a mouthful...
Let's try this:  given an angle $\theta$, draw a line through the origin that makes an angle of $\theta$ with the $x$-axis (measuring anti-clockwise).  That line will cross the unit circle somewhere.  The place where the line and circle cross is a point with coordinates $(x_{\theta},y_{\theta})$ (where the subscripts indicate that the coordinates have something to do with $\theta$).  By definition, $\sin(\theta) = y_{\theta}$.
So... suppose that we know that $\sin(\theta) = a$.  How many lines through the origin are there that intersect the unit circle at a height of $a$?  Each of those points corresponds to a solution of $\sin(\theta) = a$.  The collection of all points of height $a$ is a horizontal line.  How many times can such a horizontal line cross the circle?
Can you figure it out now?
