Where does $(\tan x)^{\sin x}=(\cot x)^{\cos x}$?

Question

So I was yet again searching throughout the homepage of Youtube to see if there were any math equations that I thought that I might be able to solve when I came across this video by SyberMath that asked the question:

$$\text{Solve: }~~~(\tan x)^{\sin x}=(\cot x)^{\cos x}$$

Which I thought that I might be able to solve. Here is my attempt at solving it:

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$$(\tan x)^{\sin x}=(\cot x)^{\cos x}$$

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$$\sin(x)\ln(\tan x)=\cos(x)\ln(\cot x)$$

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$$\tan(x)\ln(\tan x)=\ln(\cot x)~~~~~~~\text{ (since }\frac{\sin(x)}{\cos(x)}=\tan(x)\text{)}$$

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$$\text{Therefore, we need to find where }(\tan x)^{\tan x}\text{ is equal to }\cot(x)$$

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$$\text{And solving for }x\text{ will get us that }x=\frac{\pi}{4}+\pi n,n\in\mathbb{Z}\text{, (}\mathbb{Z}\text{ being the set of integers)}$$

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And, plugging this into Desmos we see that we are correct!

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$$\mathbf{\text{My question}}$$

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Is my solution correct, or is there anything that I could do to attain the correct solution more easily?

Answer 1

$(\tan x)^{\sin x}=(\cot x)^{\cos x}$ and $\cot x=\frac{1}{\tan x}$, we get

$$(\tan x)^{\sin x}=(\tan x)^{-\cos x}\Rightarrow (\tan x)^{\sin x+\cos x}=1$$

Case.(1) $$\tan x=1\Rightarrow x=\frac{\pi}4+n\pi, ~~n\in \mathbb{Z}$$

Case.(2）$$\sin x+\cos x=0\Rightarrow \sin\left(x+\frac{\pi}4\right)=0\Rightarrow x=-\frac{\pi}4+n\pi, ~~n\in \mathbb{Z}$$

The solution is:

$$ x=\pm\frac{\pi}4+n\pi, ~~n\in \mathbb{Z}$$

Answer 2

For the sake of brevity, let $s = \sin x$ and $c = \cos x$. Then $\tan x = \frac{s}{c}$ and $\cot x = \frac{c}{s}$, so:

$$(\frac{s}{c})^s=(\frac{c}{s})^{c}$$

Taking the logarithm of each side gives:

$$s(\ln s - \ln c)=c(\ln c - \ln s)$$

Which can be rearranged to:

$$(s + c)(\ln s - \ln c) = 0$$

If $s + c = 0$, i.e., $\sin x + \cos x = 0$, then $x \in \{ \frac{3\pi}{4}, \frac{7\pi}{4} \} + 2\pi n, n\in\mathbb{Z}$.

If $\ln s - \ln c = 0$, then $\sin x = \cos x$, so $x \in \{ \frac{\pi}{4}, \frac{5\pi}{4} \} + 2\pi n, n\in\mathbb{Z}$.

Combining these cases gives $x \in \{ \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \} + 2\pi n, n\in\mathbb{Z}$.

In other words, $x$ can be any odd multiple of $\frac{\pi}{4}$.

$$\boxed{x = \frac{\pi}{4}(2n+1), n \in \mathbb{Z}}$$

Answer 3

As a variant of the method given by Dan, again letting $s = \sin x, c = \cos x$, we have:

$$\left( \frac s c \right)^s = \left( \frac c s \right)^c \iff \frac{s^s}{c^s} = \frac{c^c}{s^c} \iff s^{s+c} = c^{s+c}$$

Rather than taking the logarithm, we've simply cross-multiplied. $a^n = b^n \iff a=b \lor n=0$. (That is, this is true when the base is the same on both sides, or when both exponentials are $1$.) This corresponds to $\sin x = \pm \cos x$, which is true at all odd multiples of $\tau/8$, as shown by others.