How to solve this equation $2\cos(\frac {x^2+x}{6})=2^x+2^{-x}$

Question

How do I solve for $x$ from this equation?

$$2\cos\frac {x^2+x}{6}=2^x+2^{-x}$$

Answer 1

$-2\le2\cos(\frac{x^2 + x}{6})\le 2$ and $2\le2^x + 2^{-x}$.

Therefore the only possible solution is when both equal $2$.

Answer 2

Get bounds for the left hand side and the right hand side separately ...

Answer 3

Let $y=2^x$, then $$ \begin{align} \ln y&=\ln2^x\\ \ln y&=x\ln 2\\ y&=e^{x\ln 2}. \end{align} $$ Consequently, $2^{-x}=e^{-x\ln 2}$ and $$ \begin{align} 2\cos\left(\frac{x^2+x}{6}\right)&=e^{x\ln 2}+e^{-x\ln 2}\\ \cos\left(\frac{x^2+x}{6}\right)&=\frac{e^{x\ln 2}+e^{-x\ln 2}}{2} \end{align} $$ Now, let $x=i\theta$, then $$ \begin{align} \cos\left(\frac{(i\theta)^2+i\theta}{6}\right)&=\frac{e^{i\theta\ln 2}+e^{-i\theta\ln 2}}{2}\\ \cos\left(\frac{-\theta^2+i\theta}{6}\right)&=\cos(\theta\ln 2)\\ \frac{-\theta^2+i\theta}{6}&=\theta\ln 2\\ \theta^2+(6\ln2-i)\theta&=0\\ \theta(\theta+6\ln2-i)&=0\\ \theta_1=0&\text{ or }\ \theta_2=i-6\ln2. \end{align} $$ Thus, $\large x_1=0$ and $\large x_2=-(1+6i\ln2)$.

Answer 4

I will give a hint. :) Looking at the question,you should be able to figure out that ${x = 0}$ is a solution. Apart from that, no solutions exist since ${2^x + 2^{-x} > 2}$. :))