jmc

For each point on the unit circle with $y$-coordinate equal to $-0.31$, there is a corresponding angle whose sine is $-0.31$. There are two such points; these are the intersections of the unit circle and the line $y=-0.31$, shown in red above. Therefore, there are $2$ values of $x$ with $0^{\circ }\le x<360^{\circ }$ such that $\sin x=-0.31$. There are also two values of $x$ such that $360^{\circ }\le x<720^{\circ }$ and $\sin x=-0.31$, and two values of $x$ such that $720^{\circ }\le x<1080^{\circ }$ and $\sin x=-0.31$.

But we're asked how many values of $x$ between $0^{\circ }$ and $990^{\circ }$ satisfy $\sin x=-0.31$. As described above, there are 4 such values from $0^{\circ }$ to $720^{\circ }$, but what about the two values between $720^{\circ }$ and $1080^{\circ }$?

We see that the points on the unit circle with $y=-0.31$ are in the third and fourth quadrants. So, the angles between $720^{\circ }$ and $1080^{\circ }$ with negative sines are between $720^{\circ }+180^{\circ }=900^{\circ }$ and $1080^{\circ }$. Moreover, the one in the third quadrant is less than $720^{\circ }+270^{\circ }=990^{\circ }$, so the angle in the fourth quadrant must be greater than $990^{\circ }$. This means that there's one value of $x$ between $720^{\circ }$ and $990^{\circ }$ such that $\sin x=-0.31$. Therefore, we have a total of $\boxed{5}$ values of $x$ such that $\sin x=-0.31$.