Trigonometric equation: $\ln(\sin x + \cos x)^{1+\sin 2x}= 2$ 
$\ln(\sin x + \cos x)^{1+\sin 2x}= 2$

I am unable to solve it. I tried this way:
$(\sin x + \cos x)^{1+\sin 2x}= e^2$
I know that:
$\sin x + \cos x \le \sqrt2 $
$1+ \sin 2x \le 2 $
I don't know how to utilise this idea in my solution. 

Let $X = \sin x + \cos x$. Since $1+\sin 2x = (\sin x + \cos x)^2$, we have
$$ X^2 \ln X = 2. $$
Then
\begin{align}
X^2 \ln X^2 = 4 \implies Y \ln Y = 4
\end{align}
where $Y = X^2$.
It follows from here that $Y = e^{W(4)}$. 
Thus $X = e^{W(4)/2}$.
Since 
$$X = \sin x + \cos x = \sqrt{2}\sin(x + \pi/4)$$
we conclude that
\begin{equation}
x = \sin^{-1}(e^{W(4)/2}/\sqrt{2}) - \pi/4.
\end{equation}
Added: However, since $e^{W(4)/2}/\sqrt{2} > 1$, there is no solution.
