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.
 A: $$\frac{\sin^3x}{1-\sin x}+\frac{\cos^3x}{1-\cos x}=-1\tag1$$
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.
A: It's purely algebraic. Your equation is equivalent to
\begin{align*}&\sin^3x+\cos^3x-\sin^3x\cos x-\sin x\cos^3x=\sin x+\cos x-\sin x \cos x-1\\
&\iff (\sin x+\cos x)(1-\sin x\cos x)-\sin x\cos x=\sin x+\cos x -\sin x\cos x-1\\
&\iff -(\sin x+\cos x)\sin x\cos x=-1\iff\sqrt2\cos\Bigl(x-\frac\pi4\Bigr)
\frac12\sin 2x=1\\&\iff\cos\Bigl(x-\frac\pi4\Bigr)\sin 2x=\sqrt 2
\end{align*}
which of course has no solution since $\sqrt2>1$.
A: $$\frac{\sin^3 x}{1-\sin x}+\frac{\cos^3 x}{1-\cos x}=-1$$
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.
A: Clearing fractions, you can re-write the equation as
$$\cos x + \sin x - \cos 3 x + \sin 3 x = 4$$
The left-hand side can only equal the right-hand side if all terms hit their maximum value of $1$ for the same $x$, but this is impossible. (Besides, $\sin x = 1$ and $\cos x = 1$ are disqualified by the original form of the equation.)
A: Note that the RHS is equal to $-\sin^2 x - \cos^2 x.$  Moving these terms to the LHS and adding to the fractions we get $$ \frac{\sin^2 x}{1-\sin x} + \frac{\cos^2 x}{1-\cos x} =0,$$ which never holds because either both summands are strictly positive or one is null and the other equals $1/2$. 
A: Here is a method.
Let $$a = \frac{\sin^3 x}{1-\sin x},$$ and $$b=\frac{\cos^3 x}{1-\cos x}.$$
We find the minimum of both $a$ and $b$ and show that $a+b > -1$. This can be accomplished by taking a derivative and comparing zeroes.
So,
$$\frac{d}{dx}a = {-\dfrac{\cos\left(x\right)\sin^{2}\left(x\right)\left(2\sin\left(x\right)-3\right)}{{\left(\sin\left(x\right)-1\right)}^{2}}}.$$
We can obtain zeroes in the numerator for $a'$, then plug them into $a$. The minimum should be $-\frac{1}{2}$, from what are the $3$ distinct roots per period. Finally, note that the minimum for $b$ is equal to the minimum of $a$, and that $a$ is out of phase to $b$. Therefore $a+b > -1$.
A: 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.
