Approximately $2.665357079271\ldots + 2k\pi$

can you describe what you did to solve $\sin(x)^{\sin(x)}=2$ and why these steps don't work to solve $\sin(x)^{\cos(x)}=2$

@Henry Thanks for the formatting, another lesson learnt! Thanks also for the result, but an exact one would be better, because I can do approximations myself. But thank you!

@Dennis Marx It's applying $\ln$ on both sides and then substitution $\sin x=e^z$ and you'll get something like $ze^z=\ln2$ which has the form of the inverse of Lambert function. You can do something similar with the equation OP asks but the thing is $\sin x=\sqrt{1-\cos^2x}$...

@Joan S. Guillamet F. Yep, exactly like that, but then had to resort to the complex sine, but that's exactly how I did it

Are you aware that those are not all the solutions of the equation $\sin(x)^{\sin(x)}=2$?

@jjagmath Yes, I know that not all of them are. There are complex numbers involved, which is why the root has several branches

Maybe the following can help: if considering the substitution $u = sin(x)$ and therefore $\frac{\mathrm{d}u}{\mathrm{d}x} = cos(x)$, then \begin{align} \sin(x)^{\cos(x)}&=2\\ \cos(x)\ln\left(\sin(x)\right) &= \ln\left(2\right)\\ \frac{\mathrm{d}u}{\mathrm{d}x}\,\ln\left(u\right) = \ln\left(2\right)\\ \ln\left(u\right)\,\mathrm{d}u = \ln\left(2\right)\,\mathrm{d}x\\ \int\ln\left(u\right)\,\mathrm{d}u = \int\ln\left(2\right)\,\mathrm{d}x\\ u\left(\ln\left(u\right)-1\right) = \ln\left(2\right)x + C\\ \sin(x)\ln\left(\sin(x)\right)-\sin(x) - \ln\left(2\right)x - C = 0\\ \end{align}

Then you should say something about it in your question. Something like "This are some of the solutions of $\sin(x)^{\sin(x)}=2$

@DennisMarx You can't treat $x$ as a variable. You can't derivate or integrate wrt $x$ (or $u$). You are solving an equation, which mean that those are very specific numbers that satisfy the equation, not arguments of a function.

@jjagmath Do you have any idea how to solve the equation?

My guess is that you won't be able to find a solution without the use of a more specialized function. The first equation is very directly equivalent to $z^z=2$ which can be solved in terms of the Lambert function, but the later can be transformed to $(1-z^2)^z=4$ and solving this will probably involve others functions. I know there are several ways to generalize the Lambert $W$ function, but I have no idea if those are enough to solve that one.

There is an approximate solution in terms of generailzed Lambert function ($2.54833$)

@ClaudeLeibovici Thank you! But of course a general form/exact would be much nicer

Wolfram does not find any closed form. This is not a proof that it doesn't exist, but I think it is very unlikely that it does exist

The generalisation used in this paper does not seem to be useful either. And I don't think there are any other generalisations (I tried to search for more but couldn't find.)

Sad, but thank you anyway!

Do these answer your question? Is there any valid complex or just real solution to $\sin(x)^{\cos(x)} = 2$? What is the value of $x$ such that: ${(\sin x)}^{\cos x} = \frac{1}{2}$ , considering $0 \le x \le \frac{\pi}{2} $