Number of real solution of $2^{\sin(x)}-2e^{-\sin(x)}=2$ is

Question

Number of real solution of $\displaystyle 2^{\sin(x)}-2e^{-\sin(x)}=2$ is

What I try :: Put $\displaystyle 2^{\sin(x)}=t$,

Then $\displaystyle t-\frac{2}{t}=2\Longrightarrow t^2-2t-2=0$

$\displaystyle t^2-2t+1=3\Longrightarrow (t-1)=\pm\sqrt{3}$

Then we have $\displaystyle t=1\pm \sqrt{3}\Longrightarrow \sin(x)=\log_2(1\pm \sqrt{3})$

So here $\log_{2}(1+\sqrt{3})>1$

No solution exists

Can someone please explain any short way to solve it.

Answer 1

The Given Approach

The solution given in the question makes a mistake in the first step. It appears that the solution implies that $$ 2^{\sin(x)} = \mathrm{e}^{-\sin(x)}, $$ which is not correct. Everything after this is based on the faulty assumption, which implies that the given approach is not going to work.

Honestly, because the equation is transcendental, it is unlikely that there is going to be a "nice" analytic approach to solving it—indeed, WolframAlpha gives numerical (complex, nonreal) solutions, which suggests (but does not prove) that there probably isn't a closed form solution in terms of elementary functions.

Rough Inequalities

One approach is to play around a bit with some rough inequalities. The first term, $2^{\sin(x)}$, oscillates between $\frac{1}{2}$ and $2$; while the second term, $2\mathrm{e}^{-\sin(x)}$, oscillates between $\frac{2}{\mathrm{e}}$ and $2\mathrm{e}$. This implies that the absolute largest that their difference can be is $$2^{\sin(x)} - 2\mathrm{e}^{-\sin(x)} \le 2 - \frac{2}{\mathrm{e}} < 2. $$ That is, the difference cannot be larger than the largest possible value of the first term, minus the smallest possible value of the second term. This immediately implies that the equation cannot have any real solutions.

A Note on Process

When I was a student, I often saw solutions like the one I gave above, and felt kind of annoyed, because it felt a little like pulling a rabbit out of a hat (this isn't a terrible example of a rabbit pull—most $\varepsilon$-$\delta$ proofs are worse—but it still feels a little disingenuous). In this case, before attempting any algebra, I wanted to get a sense of whether I should expect any solutions.

The first thing I did, then, was see if a computer algebra system would give me something nice. In this case, WolframAlpha does find a couple of solutions, but they are (a) complex, nonreal, and (b) numerical solutions, not analytic solutions. This suggests that "nice" solutions don't exist, and very strongly suggests that no real solutions exist at all.

Next, I plotted the $$ 2^{\sin(x)} - 2\mathrm{e}^{-\sin(x)} - 2$$ on GeoGebra:

The picture very strongly implies that there are no possible solutions. It isn't obvious how to show that this is the case, but I now have two forms of evidence which suggest that I should be looking to show that no solutions exist, and that I shouldn't bother to actually find solutions.

My next step was to plot the two functions on the left-hand side:

The term $2^{\sin(x)}$ is in blue, and $2\mathrm{e}^{-\sin(x)}$ is in red. What I notice in this picture is that the largest the difference can be corresponds to a point where $2^{\sin(x)}$ is maximized, and $2\mathrm{e}^{-\sin(x)}$ is minimized. This suggested that I should look at the difference between these two terms, which led to the solution I gave.

Answer 2

$2^y-\dfrac{2}{e^y}=2,$ where $-1\le y \le 1$

$2^y \le 2$

The second term $\dfrac{2}{e^y}$ is positive.

The left hand side is always less than $2$. No solution.

Answer 3

Substituting $t=2^{\sin(x)}$ is a good place to start. If you do that, observe the following: First, because $\sin(x)$ has range $[-1,1]$, we can restrict our attention to $t\in[\frac12,2]$. Secondly, $$ e^{-\sin(x)} =\left( e^{\ln 2 \sin(x)}\right)^{\frac1{-\ln 2}}= t^{-1/\ln 2} $$ Thus any solution to the original equation will correspond to a solution for $t\in[\frac12,2]$ to the equation $$ t -2t^{-1/\ln 2} = 2 $$ This has no solution, which we can show by observing that $t -2t^{-1/\ln 2}$ is increasing on $[0,\infty]$. (This may be shown by taking the derivative, or observing that the $t$ term is clearly increasing and the $-2t^{-1/\ln 2}$ is a composition of two decreasing functions and therefore is increasing.) Thus, we have for all $x\in \mathbb{R}$, $$ 2^{\sin x}-2e^{-\sin x} = t -2t^{-1/\ln 2} \le 2-2\cdot 2^{-1/\ln 2} < 2 $$