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.

$$\frac{\sin^3x}{1-\sin x}+\frac{\cos^3x}{1-\cos x}=-1\tag1$$
Let $f(x)$ be the LHS of $(1)$. Then, by AM-GM inequality, we have
$$\begin{align}f(x)&=\frac{(1-\sin x)(-\sin^2x-\sin x-1)+1}{1-\sin x}+\frac{(1-\cos x)(-\cos^2x-\cos x-1)+1}{1\cos x}\\&=-\sin^2x-\sin x-1+\frac{1}{1-\sin x}-\cos^2x-\cos x-1+\frac{1}{1-\cos x}\\&=-1-\sin x-1+\frac{1}{1-\sin x}-\cos x-1+\frac{1}{1-\cos x}\\&=(1-\sin x)+\frac{1}{1-\sin x}+(1-\cos x)+\frac{1}{1-\cos x}-5\\&\ge 2\sqrt{(1-\sin x)\cdot\frac{1}{1-\sin x}}+2\sqrt{(1-\cos x)\cdot\frac{1}{1-\cos x}}-5\\&=2+2-5\\&=-1\end{align}$$
Thus, we have $f(x)\ge -1$.
The equality is attained when $1-\sin x=\frac{1}{1-\sin x}$ and $1-\cos x=\frac{1}{1-\cos x}$, i.e. when $\sin x=\cos x=0$, but there is no such $x$. So, we have $f(x)\gt -1$.
Hence, $(1)$ has no solution.
