Dennis Marx

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}

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