How would I solve this trigonometric equation $\ln[(\sin(x)+\cos(x))^{1+\sin(2x)}]=2$? How would I simplify this? I can't understand which log property to use to simplify this to get the solution of this trigonometric equation

$$\ln[(\sin(x)+\cos(x))^{1+\sin(2x)}]=2$$

I know that $1+\sin(2x) = (\sin(x)+\cos(x))^2$, but I cant understand which property of logarithms to apply to get the solution here.
The equation can be written as:

$$t^{t^2}=e^2$$

where $t = \sin x+\cos x$. The function $f:(0,\sqrt{2}]\to \mathbb{R}, f(t)=t^{t^2}$, has one local minimum at $t=\frac{1}{\sqrt{e}}$, so:

$$f(t)\leq \max \{\lim_{x\to 0^+} f(x),f(\sqrt{2})\}=f(\sqrt{2})=2&lt;e^2$$

because $\lim\limits_{x\to 0^+} f(x)=1$. The equation does not have any solution.
If we use ln a^b = b ln a then we get (sin x + cos x)^2 ln (sin x + cosx) = 2.............but max value of sin x + cos x is root 2 and if we put that we get ln 2 so I dont think any real solution exists. If you want to solve for unreal solution put cos x = (e^ix + e^-ix)/2 and sin x = (e^ix - e^-ix)/2i.