Can we solve $A+D\sin^{2}x=B\sin x+C\cos x$ without having to solve a quartic polynomial? Suppose the following equation
$$
A+D\sin^{2}x=B\sin x+C\cos x,
$$
where $A,B,C,D\in\mathbb{R}$ are the real constants. Initially, I tried to find its solution from a simple substitution
\begin{align*}
A-B\sin x+D\sin^{2}x & =\pm C\sqrt{1-\sin^{2}x},
\end{align*}
that after $t=\sin x$ leads to the following quartic equation
$$
(A^{2}-C^{2})-2ABt+(B^{2}+2AD+C^{2})t^{2}-2BDt^{3}+D^{2}t^{4}=0.
$$
The Weierstrass substitution, where $t=\tan\frac{x}{2}$, and
$$
\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}=\frac{2\sin\frac{x}{2}}{\cos\frac{x}{2}}\cos^{2}\frac{x}{2}=\frac{2\tan\frac{x}{2}}{\frac{1}{\cos^{2}\frac{x}{2}}}=\frac{2\tan\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}=\frac{2t}{1+t^{2}},
$$
and
$$
\cos x=\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}=\left(1-\frac{\sin^{2}\frac{x}{2}}{\cos^{2}\frac{x}{2}}\right)\cos^{2}\frac{x}{2}=\frac{1-\tan^{2}\frac{x}{2}}{\frac{1}{\cos^{2}\frac{x}{2}}}=\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}=\frac{1-t^{2}}{1+t^{2}},
$$
leads to
$$
A+D\frac{4t^{2}}{(1+t^{2})^{2}}=B\frac{2t}{1+t^{2}}+C\frac{1-t^{2}}{1+t^{2}},
$$
which is also the quartic equation for $t$
$$
(A+C)t^{4}-2Bt^{3}+2(A+2D)t^{2}-2Bt+A-C=0,
$$
$$
(A+C)t^{4}-2Bt(t^{2}+1)+2(A+2D)t^{2}+A-C=0.
$$
Is there any better substitution avoiding the transformation of the trigonometric identity to the quartic equation for $t$?
Thanks for your help.

If you set $X=\cos x$ and $Y=\sin x$, the equation becomes geometrically finding the intersection of $DY^2-CX-BY-A=0$ (a parabola when $D\ne0$), with the circle $X^2+Y^2=1$, which generally has four solutions. So, no: you can't reduce the resolvent equation from degree $4$ unless there is some particular relation between the coefficients.
If $D=0$ the problem becomes intersecting a line (provided one among $B$ and $C$ is nonzero) with a circle and indeed can be reduced to a quadratic equation.
It's no better if you use $A=A\cos^2x+A\sin^2x$, because the conic becomes
$$
AX^2+(A-D)Y^2+CX+BY=0
$$
which is either an ellipse, for $A(A-D)>0$, or a hyperbola, for $A(A-D)<0$.
Another way to see it is setting $t=\tan(x/2)$, which transforms the equation into
$$
A+\frac{4Dt^2}{(1+t^2)^2}=\frac{2Bt}{1+t^2}+\frac{C(1-t^2)}{1+t^2}
$$
which is what you did. As you see, this generally is a quartic:
$$
(A+C)t^4-2Bt^3+(2A+4D)t^2-2Bt-C=0
$$
One gets a symmetry when $B=0$, because in this case the parabola is symmetric with respect to the $X$-axis: as you see, the quartic becomes biquadratic.
