We have,
$$
\Rightarrow \quad \begin{aligned}
\sin \theta+\sin ^2 \theta & =1 \Rightarrow \sin \theta=1-\sin ^2 \theta \\
\sin \theta & =\cos ^2 \theta
\end{aligned}
$$
We know that,
$$
\begin{aligned}
&\left(\sin ^2 \theta+\cos ^2 \theta\right)^3-1=0 \\
& \Rightarrow \quad\left(\cos ^4 \theta+\cos ^2 \theta\right)^3-1=0 \\
& \Rightarrow\left(\cos ^{12} \theta+3 \cos ^{10} \theta+3 \cos ^8 \theta+\cos ^6 \theta\right)-1=0 \\
& \Rightarrow \cos ^{22} \theta+3 \cos ^{20} \theta+3 \cos ^8 \theta+\cos ^6 \theta-1=0
\end{aligned}
$$
On comparing, we get
$$
\begin{aligned}
a & =3, b=3, c=1, d=-1 \\
\therefore \quad a c+b d & =3 \times 1+3 \times-1=3-3=0
\end{aligned}
$$