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. 

HINT
$$\ln(\sin x + \cos x)^{1+\sin 2x}= 2\iff\ln(\sin x + \cos x)^{1+\sin 2x}= \ln e^2\iff (\sin x + \cos x)^{1+\sin 2x}=e^2$$
but
$$(\sin x + \cos x)^{1+\sin 2x}=(\sin x + \cos x)^{(\sin x + \cos x)^2}\leq\sqrt2^{2}=2$$
Note
$$(\sin x + \cos x)^2=1+\sin 2x\le2\implies \sin x + \cos x \leq\sqrt 2$$
