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.

The minimum of $\frac{\sin^3}{1-\sin{x}}$ is $-\frac12$, so, except for $\sin{x}=-1$, $\frac{\sin^3}{1-\sin{x}}\gt-\frac12$.
The minimum of $\frac{\cos^3}{1-\cos{x}}$ is $-\frac12$, so $\frac{\cos^3}{1-\cos{x}}\ge-\frac12$.
So, except for $\sin{x}=-1$, $\frac{\sin^3}{1-\sin{x}}+\frac{\cos^3}{1-\cos{x}}\gt-1$.
If $\sin{x}=-1$, then $\cos{x}=0$, so $\frac{\sin^3}{1-\sin{x}}+\frac{\cos^3}{1-\cos{x}}=-\frac12+0$.
So, the equation has no solution.
