= Solution
For any distinct nodes, the <polynomial> $q(x)=\prod_{j=1}^\nu(x-c_j)^2$ has degree $2\nu$, is nonnegative, and vanishes at every node. The <quadrature rule> sum for $q$ is zero, but $\int_0^\pi wq\,dx>0$ because the weight is positive and $q$ is positive except at finitely many points. Therefore \b[no such rule is exact for all polynomials of degree at most $2\nu$], irrespective of the chosen weights.
For the sine weight, its first four <moments> are $2,\pi,\pi^2-4,\pi^3-6\pi$. A monic degree-two <orthogonal polynomial> is
$$
p_2(x)=x^2-\pi x+2,
$$
since direct use of these <moments> gives $\int p_2\sin x\,dx=\int xp_2\sin x\,dx=0$. Its roots lie in $(0,\pi)$ and are
$$
\boxed{c_{1,2}=\frac\pi2\mp\sqrt{\frac{\pi^2}{4}-2},\qquad b_1=b_2=1.}
$$
The weights follow by requiring exactness for $1$ and $x$: $b_1+b_2=2$ and $b_1c_1+b_2c_2=\pi$, and the symmetric nodes imply both weights are one. To prove cubic exactness, divide any cubic as $f=ap_2+r$, with $a,r$ of degree at most one. The <integral> of $ap_2$ is zero by <orthogonality>, and its nodal values vanish; the remaining linear <polynomial> is integrated exactly. This is the <two-node Gaussian quadrature with sine weight>; the argument proves its <Gaussian quadrature> property without assuming a general quadrature theorem.
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