What is sum of the roots of the equation $(\sin x+\cos x)^{(1+\sin 2x)}=2\quad$ where $x\in[-2\pi,4\pi]$ ?
We have $1+\sin 2x=\sin^2 x+\cos^2x+2\sin x\cos x=(\sin x+\cos x)^2$. so the equation is $$(\sin x+\cos x)^{(\sin x+\cos x)^2}=2$$ By taking $\sin x+\cos x=u$ we have $$u^{u^2}=2$$ By try and error I realized that $u=\pm\sqrt2$ are the answers , but I don't know how to solve the above equation mathematically to find out whether it has other answers or not.
Here's how to solve anallitically $x^{x^a} = b$. Raising to the power of $a$ we have $(x^a)^{x^a} = b^a$. Let $y=x^a$ and $c=b^a$ so we need to solve $y^y=c$. This can be solved in terms of the Lambert function like this: $y=e^{W(\log c)}$.
One need to be careful since $W$ has two real branches (as the square root, but this two branches have different domains, see the link).
In the case of your equation $u^{u^2}=2$, this solution gives, after some simplification, $u = \pm\sqrt{2}$.
$$x=n^{\frac{1}{n}}$$ satisfies the equations $$x^n=n,~~x^{x^n}= n,~~\ldots,~~x^{x^{x^{\cdots^{x^n}}}}=n$$
So your case is really just a special case, with $n=2$. Therefore $$x=2^\frac{1}{2}=\sqrt 2$$ is a solution.
I learnt this trick from this video; blackpenredpen's channel has many more awesome tricks, I really recommend it.
hint
$$\sin(x)+\cos(x)=\sqrt{2}\sin(x+\frac{\pi}{4})$$
For $ u>1$, put $$f(u)=u^2\ln(u)$$ $$f'(u)=u(2\ln(u)+1)>0$$ $ f $ is strictly increasing at $(1,+\infty)$ and $ f(\sqrt{2})=\ln(2)$ so, $ \sqrt{2} $ is the only root of $ f(x)=\ln(2) $ at $ (1,+\infty)$.
Let us try to solve $u^{u^2}=2$ for $|u| \leq \sqrt{2}$.
Case 1: $u$ is positive.
Then, we can take log on bot sides and the equation becomes $$ u^2 \ln(u)= \ln (2) $$ This gives $\ln(u) >0$ and hence $u >1$.
Note here that the function $$ f: (1, \infty) \to \mathbb R \,;\, f(u)=u^2 \ln(u) $$ is the product of two strictly increasing positive functions and hence strictly increasing. Therefore, as $f(\sqrt{2})=\ln(2)$ the equation $$ f(u)=u^2 \ln(u) $$ has unique solution $u =\sqrt{2}$.
Case 2: $u$ is negative.
Set $v=-u$. Then $u^2=-v^2$ and $|u|=v$.
Then $u^{u^2}=2$ gives $$ |u^{u^2}|=2 \Rightarrow |u|^{u^2}=2 \Rightarrow v^{v^2}=2 $$ Therefore, if $u$ is a negative solution, $v=-u$ is a positive solution to $v^{v^2}=2$ and hence $v=\sqrt{2}$ by Case 1.
This shows that the only potential negative solution is $u=-\sqrt{-2}$.
You need to check that this works.
Let $\cos x+\sin x=\sqrt2\cos y$ where $y=x-\dfrac\pi4$
$$\implies2=(\sqrt2\cos y)^{2\cos^2y}=2^{\cos^2y}(\cos^2y)^{\cos^2y}$$
Now $1\le2^{\cos^2y}\le2^1$ and $0<(\cos^2y)^{\cos^2y}\le1$
$$\implies0<2^{\cos^2y}(\cos^2y)^{\cos^2y}\le2$$
The equality will occur if $2^{\cos^2y}=2, (\cos^2y)^{\cos^2y}=1$ which is possible iff $$\cos^2y=1\iff\sin y=0$$
