Solve for $x$ where $\sin(x)^{\cos(x)}$= $2$

Question

I found this question in a mathematics book (some kind of challenge questions book). I was just trying to solve this question for fun. At the moment I am a bit confused. So by now, I have tried two things 1st of all is to sub $cos(x)$ as

$\frac{1}{\pm \sqrt{1+\tan^2(x)}}$ and $\pm \sqrt{\sin^2(x)-1}$.

However, I haven't made any progress. I graphed both equation son desmos and Graph of both sides of the equation. I also got this solution from a math computation website solution. Any help will be highly appreciated.

Answer 1

First approach

Plotting or by inspection, the first solution is close to $\frac {5 \pi}6$.

Now, consider $$f(x)=\cos(x)\log(\sin(x))-\log(2)$$ and expand as a Taylor series $$f(x)=\frac{1}{2} \left(\sqrt{3}-2\right) \log (2)+\frac{1}{2} \left(x-\frac{5 \pi }{6}\right) (3+\log (2))-$$ $$\frac{1}{4} \left(x-\frac{5 \pi }{6}\right)^2 \left(\sqrt{3} (\log (2)-6)\right)+\left(x-\frac{5 \pi }{6}\right)^3 \left(\frac{9}{4}-\frac{\log (2)}{12}\right)+$$ $$O\left(\left(x-\frac{5 \pi }{6}\right)^4\right)$$

Using series reversion $$x=\frac{5 \pi }{6}+t+\frac{\sqrt{3} (\log (2)-6)}{6+\log (4)}t^2+\frac{ (243+\log (4) (\log (32)-66))}{6 (3+\log (2))^2}t^3+O\left(t^{4}\right)$$ where $t=\frac{2 f(x)+(2-\sqrt{3}) \log (2)}{3+\log (2)}$.

Since you want $f(x)=0$, then the approximation $x \approx 2.6653795$ while the "exact" solution, given by Newton method, is $x=2.6653571$

Second approach

Using the tangent half-angle substitution $x=2\tan^{-1}(t)$, you need to find the zero of function $$g(t)=\left(\frac{2 t}{t^2+1}\right)^{\frac{1-t^2}{1+t^2}}-2$$ Plotting, the solution is close to $t=4$ and the function is very close to linearity. Using Newton method, the iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 4.000000000 \\ 1 & 4.120370127 \\ 2 & 4.119928024 \\ 3 & 4.119928018 \end{array} \right)$$ Back to $x$, $x=2.6653571$.

Third approach

The derivative of $f(x)$ cancels at $x=\frac \pi 2$ and the second derivative test shows that it is a local minimum. Expand as a series $$f(x)=\frac{1}{2} \left(x-\frac{\pi }{2}\right)^3+\frac{1}{80} \left(x-\frac{\pi }{2}\right)^7+O\left(\left(x-\frac{\pi }{2}\right)^9\right)-\log(2)$$ Using the first term only, a first estimate $$x_0=\frac{\pi }{2}+\sqrt[3]{2 \log (2)}=2.68582$$ Plug this is the second term $$0=\frac{1}{2} \left(x-\frac{\pi }{2}\right)^3+\frac{1}{80} \log ^{\frac{7}{3}}(4)-\log(2)$$ $$x_1=\frac{\pi }{2}+\sqrt[3]{\log (4)-\frac{1}{40} \log ^{\frac{7}{3}}(4)}=2.67127$$

Answer 2

We have

$$\sin(x)^{\cos(x)} = 2$$

We'd want to take $\cos(x) = \sqrt{1 - [\sin(x)]^2}$, but the solutions can only happen when $\cos(x)$ is negative, because there is no other way to get a value larger than 1, given that $\sin(x)$ is never larger than 1. Hence we must negate the square root, giving

$$\sin(x)^{-\sqrt{1 - [\sin(x)]^2}} = 2$$

Now take $u = \sin(x)$, so

$$u^{-\sqrt{1 - u^2}} = 2$$

Taking the log of both sides gives

$$-\sqrt{1 - u^2} \ln(u) = \ln(2)$$

Take $u = e^{v/2}$,

$$-\sqrt{1 - e^v} \frac{v}{2} = \ln(2)$$

so

$$-\sqrt{1 - e^v} v = 2 \ln(2)$$ $$(1 - e^v) v^2 = 4 [\ln(2)]^2$$ $$1 - e^v = 4 \frac{[\ln(2)]^2}{v^2}$$ $$e^v = 1 - 4 \frac{[\ln(2)]^2}{v^2} = \frac{v^2 - 4 [\ln(2)]^2}{v^2}$$

which can be factored and rearranged to

$$e^v = \frac{(v - 2 \ln(2))(v + 2 \ln(2))}{v^2}$$

But, I think, this is either as far or not as far as it gets when it comes to solving the equation, and the reason I say this is this is a special case of general equations of the type

$$e^v = \frac{(v - a)(v - b)}{(v - c)(v - d)}$$

which have been examined before:

https://arxiv.org/pdf/1408.3999.pdf

as a generalization of the Lambert W-function. Thus, there is likely to be no solution either using elementary functions or even the standard Lambert W-function as an addition. In Mezo's notation, though, we could say

$$v = W\left(\begin{matrix}0 && 0\\2 \ln(2) && -2 \ln(2)\end{matrix}; 1\right)$$

hence

$$x = \sin\left(e^{\frac{1}{2} W\left(\begin{matrix}0 && 0\\2 \ln(2) && -2 \ln(2)\end{matrix}; 1\right)}\right).$$

But I suspect hardly any are familiar or have a good intuition for the behavior of Mezo's generalized Lambert functions, so this may not be helpful or insightful. Generally speaking these equations though will not have "nice" answers to them anyways.