Prove that the trigonometric equation has no solution Prove that the trigonometric equation $$\frac{\sin^3 x}{1-\sin x}+\frac{\cos^3 x}{1-\cos x}=-1$$ has no solution. I tried applying $T2's$ lemma to contradict but could only do so for the first and third quadrant values if $x$. There must be some good proof without the restrictions in the values of $x$. Thanks.

$$\frac{\sin^3 x}{1-\sin x}+\frac{\cos^3 x}{1-\cos x}=-1$$
Obviously $\sin x\neq 1$ and $\cos x\neq 1$.
$$\sin^3 x(1-\cos x)+\cos^3 x(1-\sin x)=-1 -\sin x\cos x+\sin x+\cos x$$
$$\sin^3 x+\cos^3 x-\sin x\cos x(\sin^2 x+\cos^2 x)=-1 -\sin x\cos x+\sin x+\cos x$$
$$\sin^3 x+\cos^3 x-\sin x-\cos x=-1 $$
$$(\sin x+\cos x)(\sin ^2 x-\sin x\cos x+\cos^2 x -1)=-1$$
$$(\sin x+\cos x)\sin x \cos x=1.  $$
It would follow that $(\sin x+\cos x)^2\sin^2 x\cos^2 x=1.$
$(1+2\sin x\cos x)\sin^2 x\cos^2 x=1$
$(1+\sin (2x))\sin ^2(2x)=4$,
which has no solution. Hope the calculations are alright.
