Exponential-Trigonometric equation

Long time, No solutions, SO this is how,

By AM-GM inequality, we can see,

$2^{x}+\dfrac{1}{2^{x}} \geq 2$

We also know, $1+ \cos x = 2 \cos ^2 \dfrac{x}{2}$

Which means, 2 = LHS=RHS

$\implies 2 \cos ^2 \dfrac{x}{2} = 2$

$\implies \cos ^2 \dfrac{x}{2} =1$

$\implies \cos \dfrac{x}{2} = 1$ [Negetive is not possible for exponentials]

$\implies x=0$ seems the only solution to the question.

:D

Thanks.

Please Upvote

We note that the RHS $=\cos x + 1 \in [0, 2]$ . That is the maximum value of $\cos x + 1$ is 2 and it occurs at $\pm 2k\pi$ , where $k$ is an integer. Now consider the LHS.

Let $f(x) = 2^x + 2^{-x}$ . Then $f'(x) = \ln 2 \left(2^x - 2^{-x}\right)$ and $f(0) = 0$ . And $f''(x) = \ln^2 \left(2^x + 2^{-x}\right)$ and $f''(0) > 0$ . This mean that $f(0) = 2$ is the minimum value of $f(x)$ , the LHS.

There is a universal minimum of the LHS of 2 when $x=0$ and simultaneously, the RHS has its maximum of 2. Therefore, there is only one solution for the equation, that is $x=\boxed{0}$ .

For any positive real number $a$ we have :

$(a-1)^{2} \geq 0 \Rightarrow a^{2}+1 \geq 2a \Rightarrow a + \frac {1}{a} \geq 2$

We know that : $2^{x} \geq 0$ for any real value of x so $2^{x} + 2^{-x} \geq 2$

And (=) is true if and only if $2^{x}$ = 1 or x=0 .

On the other hand , f(x) = cos(x)+1 has maximum value at : x=0 , f(0) = 1+1 = 2

After that $f(x) \leq 2$

So x=0 is a unique real solution