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.

  • $\begingroup$ $\begin{align*} A-B\sin x+D\sin^{2}x & = \pm C\sqrt{1-\sin^{2}x} \end{align*} \not = C\sqrt{1-\sin^{2}x}$ $\endgroup$ – S.H.W Aug 5 '18 at 15:24
  • $\begingroup$ @ S.H.W: thanks, corrected $\endgroup$ – justik Aug 5 '18 at 17:55
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    $\begingroup$ Note that $A+B\sin^2 x$ can be written as $P + Q\cos 2x$ for appropriate $P$ and $Q$; also, $B \sin x + C \cos x$ can be written as $M \sin(x + N)$ for appropriate $M$ and $N$. Considering how a "$\cos 2x$" wave and a "$\sin(x+N)"$ wave look, it seems reasonable to expect/fear that —barring some special relationships among the coefficients— the curves will meet in four asymmetric points. $\endgroup$ – Blue Aug 5 '18 at 21:01
  • $\begingroup$ @ Blue: Thanks for the explanation, you are right... $\endgroup$ – justik Aug 6 '18 at 10:23

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.

  • $\begingroup$ @ egreg: Thanks for the detailed explanation. I also tried $A^2=A^2cos^2x+A^2sin^2x$ with reordering the terms, but without success. Hence, there is no way to simplify the equation :-( $\endgroup$ – justik Aug 5 '18 at 19:21

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