Finding sum of the roots of $(\sin x+\cos x)^{(1+\sin 2x)}=2$

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.

• I think that the problem is ill posed, when $x \in (\frac{3\pi}{4}, \frac{7 \pi}{4})$ the left hand side is a negative number to a real exponent, what does this mean? Apr 20 '21 at 23:38

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, $$u = \pm\sqrt{2}$$.

In general:

$$x=n^{\frac{1}{n}}$$ 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 $$(1,+\infty)$$.

• Thanks, but my real issue is solving $u^{u^2}=2$ mathematically. Apr 20 '21 at 22:45
• @Soheil Ok, i edited it. Apr 20 '21 at 22:48

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 $$\cos^2y=1\iff\sin y=0$$