What is the value of $x$ such that: ${(\sin x)}^{\cos x} = \frac{1}{2}$ , considering $0 \le x \le \frac{\pi}{2} $

Question

Doing $\sin x = u$ and $\cos x = v$ , we have $u^2 + v^2 = 1$ and the equation: $u^v = \frac{1}{2}$

Then: $(u^v)^2 = \frac{1}{4}$ , or better: $(u^2)^v = \frac{1}{4}$. Replacing $u^2$, we have:

$(1 - v^2)^v = \frac{1}{4}$ , or yet: $(1 +v)^v(1 -v)^v = \frac{1}{4}$

Now how to solve this last equation? Would we have any success if we apply Lambert W Function?

Answer 1

It would be more than difficult to find a closed form even using special functions for the zero of function $$f(x)=\sin(x)^{\cos(x)}-\frac 12 \tag 1$$

If you compose Taylor series around $x=0$

$$f(x)=-\frac 12+x-\frac{1}{6} x^3 (3 \log (x)+1)+O\left(x^5\right)\tag 2$$ So, a very first estimate is $x_0=\frac 12$.

Make one iteration of Newton method for $(2)$ to obtain $$x_1=\frac 12-\frac{3\log (2)-1}{18 (2+\log (2))}=0.477733$$ while the "exact" solution is $x=0.476236$.

Adding one term to the expansion given in $(2)$ would give instead $x_1=0.476128$

Answer 2

Is there any valid complex or just real solution to $\sin(x)^{\cos(x)} = 2$?

has a series solution for $$\sin(x)^{\cos(x)}=2\mathop\iff^{x=\cos^{-1}(-\sqrt y)}y=1-4^{-\frac1{\sqrt y}}$$

so the solution to $\sin(x)^{\cos(x)}=\frac12$ is $x=\cos^{-1}(\sqrt y)$. Thus, we use the $_1\text F_2$ function from the blockquote and therefore:

$$\bbox[border:2px dashed blue]{\begin{align}\sin(x)^{\cos(x)}=\frac12\implies x= \cos^{-1}\left( 1+\sum_{n=1}^\infty(-1)^n\binom{\frac12}n\,_1\text F_2\left(n-\frac12;\frac12,\frac12;(\ln(2)n)^2\right)+\ln(2)\,_1\text F_2\left(n;1,\frac32;(\ln(2)n)^2\right)\right)\end{align}}$$

shown here:

Answer 3

$$\sin(x)^{\cos(x)}=\frac{1}{2}$$ $$e^{\ln(-(\frac{1}{2}i)(e^{ix}-e^{-ix}))(\frac{1}{2}e^{ix}+\frac{1}{2}e^{-ix})}=1/2$$ $$\ln(-(\frac{1}{2}i)(e^{ix}-e^{-ix}))(\frac{1}{2}e^{ix}+\frac{1}{2}e^{-ix})=-\ln(2)$$

Looks too complicated.
$\ $

$$\sin(x)^{\cos(x)}=\frac{1}{2}\tag{1}$$ $\sin(x)\to\sqrt{1-\cos(x)}$: $$\sqrt{1-\cos(x)^2}^{\ \cos(x)}=\frac{1}{2}$$ $$\left(\sqrt{1-\cos(x)^2}^{\ \cos(x)}\right)^2=\frac{1}{4}$$ $$(1-\cos(x)^2)^{\cos(x)}=\frac{1}{4}$$ $\cos(x)\to t$: $$(1-t^2)^t=\frac{1}{4}$$ $$e^{\ln(1-t^2)t}=\frac{1}{4}$$ $$\ln(e^{\ln(1-t^2)t})=-2\ln(2)$$ $$\ln(1-t^2)t=-2\ln(2)$$ $t\to\sqrt{1-e^u}$: $$\ln(e^u)\sqrt{1-e^u}=-2\ln(2)$$ $$u\sqrt{1-e^u}=-2\ln(2)$$ $$u^2(1-e^u)=4\ln(2)^2$$ $$u^2-u^2e^u=4\ln(2)^2$$ $$-u^2e^u=4\ln(2)^2-u^2$$ $$\frac{-u^2}{4\ln(2)^2-u^2}e^u=1$$ $$\frac{u^2}{u^2-4\ln(2)^2}e^u=1\tag{2}$$

We see, this equation cannot be solved in terms of Lambert W, but in terms of Generalized Lambert W:

$$\frac{u^2}{(u-2\ln(2))(u+2\ln(2))}e^u=1$$ $$\frac{(u-0)(u-0)}{(u-2\ln(2))(u-(-2\ln(2)))}e^u=1$$ $$u=W\left(^{\ \ \ \ \ \ 0,\ \ \ \ \ \ 0}_{2\ln(2),-2\ln(2)};1\right)$$

Back substitution of $u$ and $t$ yields candidates for $x$. Because the square function isn't injective, squaring isn't an equivalence operation. We made two squaring steps. Therefore the solution set of equation (1) is a proper or improper subset of the solution set of equation (2).

So we have a closed form for $x$, and the series representations of Generalized Lambert W give some hints for calculating $x$.

May be there are simpler solutions.

[Mező 2017] Mező, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553

[Mező/Baricz 2017] Mező, I.; Baricz, Á.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)

[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018