<s>[INST] 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? 
 [/INST] 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, $x_1,180^\circ- x_1.$
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