Answer: A

Solution:

Since $\sin x\ne0$ on $(0,\pi),$ divide by $\sin x:$ $$\begin{gathered} a(1+2\cos x)=4\cos^2 x-1 \\ =(2\cos x-1)(2\cos x+1). \end{gathered}$$

When $\cos x=-\tfrac12$ (that is, $x=\tfrac{2\pi}{3}$) both sides vanish, so this is a solution for every $a.$ Otherwise we may cancel $1+2\cos x$ to get $a=2\cos x-1,$ i.e. $\cos x=\dfrac{a+1}{2}.$

This yields a second solution in $(0,\pi)$ exactly when $-1\lt\dfrac{a+1}{2}\lt1,$ that is $a\in(-3,1),$ and it is distinct from $x=\tfrac{2\pi}{3}$ unless $a=-2.$

So more than one solution occurs for $a\in(-3,-2)\cup(-2,1),$ giving $p+q+r=-3-2+1=-4.$

Thus, the correct answer is A.