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I'm trying to solve it by log two parts:

$$ \log_2(2^{x}/2^{-x}) = \log_2(2*\cos(x/5)) $$ Simplifying: $$ 2x - 1 = \log_2(\cos(x/5)) $$

And I'm stuck here, I think we could solve it by investigating functions(linear and logarithmic) and trying to find border value. Could you give the right direction?

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    $\begingroup$ Use \cos, \log_2, etc. and... no, $\log_2(x-y)\neq\log_2 x-\log_2 y$. $\endgroup$
    – metamorphy
    Feb 21 '21 at 7:56
  • $\begingroup$ @metamorphy, rookie mistake. Thanks! $\endgroup$
    – funnydman
    Feb 21 '21 at 8:13
  • $\begingroup$ This is a transcendental equation, ie no solutions exist that we can find in a closed form. $\endgroup$ Feb 21 '21 at 9:59
  • $\begingroup$ @A-LevelStudent, see a solution $\endgroup$
    – funnydman
    Feb 25 '21 at 7:42
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Consider that you look for the zero of function $$f(x)=2^{x} - 2^{-x} - 2\cos \left(\frac{x}{5}\right)$$ By inspection, you know that the solution is somewhere between $x=1.0$ and $x=1.5$ and the function is quite close to a straight line (this is good for root-finding methods).

So, be as lazy as I am and start Newton methods with $x_0=0$. It will give the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.0000000 \\ 1 & 1.4426950 \\ 2 & 1.2504695 \\ 3 & 1.2401243 \\ 4 & 1.2400975 \end{array} \right)$$

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We have the equation $$2^x+2^{-x}=2\cos\frac{x}{5}$$ and we want to either find the sum of the solutions or the solutions themselves. If we are only concerned with real solutions, then due to your excellent use of inequalities we have the only solution: $x=0$, and we are answering the second problem.

Alternatively, if the question also wants you to consider complex roots, then the second problem is not applicable, as there are multiple solutions, and in fact $0$ is the answer to the first problem. I will demonstrate how to find all these complex solutions below, and also very simply why $0$ is the answer to the first problem.


To find the solutions to this equation:

First observe that $2^x=e^{x\ln2}$ and similarly, we have $2^{-x}=e^{-x\ln2}$. Hence our equation is equivalent to $$2\left(\frac{e^{x\ln2}+e^{-x\ln2}}{2}\right)=2\cos\frac{x}{5}$$ But if you recall the definition of $\cosh A$ then you should see that actually our equation is equivalent to $$2\cosh(x\ln2)=2\cos\frac{x}{5}\iff \cosh(x\ln2)=\cos\frac{x}{5}$$ However, we also have a useful identity $$\cosh(A)\equiv \cos(iA)$$ Thus, we actually have $$\cos{ix\ln2}=\cos\frac{x}{5}\implies ix\ln2=\pm\frac{x}{5}+2n\pi\implies x=\frac{2n\pi}{i\ln2\mp 0.2}=\frac{10n\pi}{5i\ln 2\mp1}$$ for integer $n$.


To answer the question:

Note that if $\alpha$ is a solution of the equation, then so is $-\alpha$. Therefore there must be an even number of roots (in fact there are infinitely many complex roots, but since every roots is paired up with another one which is equal to the negative of that root, loosely speaking we have an even number of roots). So if there are $2n$ roots, $$\alpha_1,-\alpha_1,\ldots,\alpha_n,-\alpha_n$$ then the sum of the roots is $$\alpha_1+(-\alpha_1)+\cdots+\alpha_n+(-\alpha_n)=0$$ as required.

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  • $\begingroup$ Thanks. It's a very good explanation! $\endgroup$
    – funnydman
    Feb 25 '21 at 13:01
  • $\begingroup$ @funnydman No problem! I'm really happy to have helped you :) $\endgroup$ Feb 25 '21 at 13:08
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Looks like there a typo in the textbook, the left part must be $2^x + 2^{-x}$. In this case, I found a solution:

Since $2^x + \frac{1}{2^{x}}\geq 2$ and $2\cos{\frac{x}{5}} \leq 2$,then

\begin{cases} 2^x + 2^{-x}=2 \\ 2\cos{\frac{x}{5}}=2 \\ \end{cases}

After solving this simple system, the answer would be $ x = 0 $.

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  • $\begingroup$ How did you use the fact $2^x+\frac{1}{2^x}\ge2$ in your solution? Weren't you dealing with $2^x-2^{-x}$ (stress on the negative sign)? $\endgroup$ Feb 25 '21 at 9:18
  • $\begingroup$ To clarify: the correct equation should be $2^x+2^{-x}=2\cos(x/5)$? $\endgroup$ Feb 25 '21 at 10:14
  • $\begingroup$ @A-LevelStudent, I thought about a typo too, but I'm not sure about this. But I know the answer from the textbook, the question is "Find the sum of the roots or the root of the equation, if it is the only one" and the answer is 0. $\endgroup$
    – funnydman
    Feb 25 '21 at 10:16
  • $\begingroup$ @A-LevelStudent, what do you think? I appreciate your attention. $\endgroup$
    – funnydman
    Feb 25 '21 at 10:19
  • $\begingroup$ I think the solution conclusively proves that there's a typo. If there wasn't a typo: $0$ isn't the answer to the first problem (see Claude's answer above) as there's only one real solution, and since there's only one real solution which is non zero, $0$ is certainly not the answer to the second problem either. I'll type an answer for what must be the correct equation; it'll add a little to your current solution. $\endgroup$ Feb 25 '21 at 10:56

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