The given equation is equivalent to $$7\sin x + 2(1-\sin^2 x)=5\implies 2\sin^2 x - 7\sin x + 3 = 0\implies  \sin x =\frac{7\pm\sqrt{(-7)^2 - 4\times 2\times 3}}{4}=3, \frac{1}{2}$$

However, we have $\sin x \le 1$, therefore we find $\sin x=\frac{1}{2}$. There will be $\boxed{2}$ solutions in the given range.