Prove that the trigonometric equation has no solution

Question

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.

Answer 1

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 $1/2$.

Answer 2

$$\frac{\sin^3x}{1-\sin x}+\frac{\cos^3x}{1-\cos x}=-1\tag1$$

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.

Answer 3

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 $\sqrt2>1$.

Answer 4

$$\frac{\sin^3 x}{1-\sin x}+\frac{\cos^3 x}{1-\cos x}=-1$$

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.

Answer 5

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.)

Answer 6

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 $a+b > -1$.

Answer 7

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.