# Is $\log_{\cos x}(1)$ defined at $x=0+2k\pi$? [duplicate]

I have an equation like this:

$\cos(x) ^ {\sin(x)} = 1$

I thought I would solve it like this:

$\cos(x) ^ {\sin(x)} = 1$
$\sin(x) = \log_{\cos(x)}(1)$
$\sin(x) = 0$
$x = 0+k\pi$

But I'm confused about the domain.
I did $\log_{\cos(x)}(1)$ which should mean $(\cos(x)>0 \wedge \cos(x)\neq1)$
i.e $x\in(-\frac{\pi}{2}+2k\pi, 0+2k\pi) \cup (0+2k\pi, \frac{\pi}{2}+2k\pi)$
but then all the solutions of $x = 0+k\pi$ are invalidated.

Is $\log_{\cos(x)}(1)$ always considered $0$ no matter the $x$?

• It's a bad idea to write $\log_{\cos(x)}$ (it's not even clearly defined), just write that $\cos(x)^{\sin(x)} = e^{\sin(x) \ln(\cos(x))}$, and all become clear – Tryss Apr 15 '15 at 22:02
• @Tryss Sorry, I deleted the comment once I realised that they're just equivalent statements, the one you wrote and my rearrangement. Definitely, writing log with a functional base is to be avoided. – nathan.j.mcdougall Apr 15 '15 at 22:19

Here is an easier way out. Note that if $x \neq n\pi$, we have $\vert \cos(x) \vert < 1$. This would mean $\vert\cos(x)\vert^{\sin(x)}<1$, since $\sin(x) \neq 0$ as $x \neq n\pi$. Hence, we only need to consider the case $x=n\pi$. We see that $x=n\pi$ gives us that $\cos(n\pi)^{\sin(n\pi)}=1$.

We should take care of the case $\cos(x) = 0$ separately: when $\cos(x) = 0$, $\sin(x) = \pm 1$, and $0^1 = 0$ while $0^{-1}$ is undefined; neither are $1$, so no solutions there.

You do have $\cos(x)^{\sin(x)} = (\pm 1)^0 = 1$ when $x = n \pi$. Otherwise, if you're interested in real solutions, you have $0 < |\cos(x)| < 1$ and either $\sin(x) > 0$ or $\sin(x) < 0$. If $\sin(x) > 0$, $\left|\cos(x)^{\sin(x)}\right| = |\cos(x)|^{\sin(x)} < 1$, and if $\sin(x) < 0$, $\left|\cos(x)^{\sin(x)}\right| = |\cos(x)|^{\sin(x)} > 1$, so no more solutions there.

However there can be complex solutions. By definition, $\cos(x)^{\sin(x)} = \exp(\sin(x) \ln(\cos(x)))$. Note that this can involve the logarithm of negative numbers: that's OK if you allow complex values. If you're not comfortable with complex numbers, don't even consider non-integer powers of negative numbers.

Now $1 = \exp(0)$, so you have solutions when $\sin(x) \ln(\cos(x)) = 0$. This would mean either $\sin(x) = 0$ (which happens when $x = n \pi$ for integer $n$) or $\ln(\cos(x)) = 0$ (which means $\cos(x) = 1$, and that's included in the previous solutions).

However, the complex logarithm is multi-valued, so there are other cases to consider.
$1 = \exp(z)$ for $z = 2 \pi i m$ for integer $m$, so you want to consider $\sin(x) \ln(\cos(x)) = 2 \pi i m$.

Now if $x = it$ is imaginary, $\cos(x) = \cosh(t)$ and $\sin(x) = i \sinh(t)$, so you're looking at $\sinh(t) \ln (\cosh(t)) = 2 \pi m$. The graph of $\sinh(t) \ln(\cosh(t))$ looks like this:

so there should be one solution with real $t$ for every $m$. It probably doesn't have a closed form, but can be obtained numerically. For $m = 1, 2, 3$ the solutions are approximately $t = 2.155448920$, $2.587822150$, $2.858727129$ respectively.

There are also lots of solutions with both real and imaginary parts nonzero, e.g. (approximately) $$\pm 2.03750688737218 \pm 1.68500078184922 i$$. Here are all the solutions that I've found using Maple in the rectangle $\{x+iy: -\pi \le x \le \pi, -5 \le y \le 5\}$:

Note that since $\cos$ and $\sin$ are periodic with period $2\pi$, this picture repeats horizontally.