Sine is Boxed

Possible values of $\lfloor \sin p\rfloor$ are $-1,0,1$ .

• Solving $\frac{5p^2 - 2p}{6p-3} = 1$ yields $5p^2 - 8p + 3 = 0$ , so that $p = 1,-\tfrac35$ . Since $\lfloor \sin p\rfloor$ is equal to $0,-1$ respectively in these cases, there are no solutions here.
• Solving $\frac{5p^2 - 2p}{6p - 3} = 0$ yields $p =0,\tfrac25$ , and $\lfloor \sin p\rfloor = 0$ for both of these values. We have found $2$ solutions so far.
• Solving $\frac{5p^2 - 2p}{6p-3} = -1$ yields $5p^2 + 4p - 3 = 0$ , so that $p = -\tfrac25 \pm \tfrac15\sqrt{19}$ . The values of $\lfloor \sin p\rfloor$ for these two numbers are $0,-1$ respectively ( $+$ first, then $-$ ), so we pick up another solution.

Thus $p = 0,\tfrac25, -\tfrac25-\tfrac15\sqrt{19}$ are the $\boxed{3}$ solutions.