No. of real solutions of the equation $2 \cos (\frac{x^2 + x}{6}) = 2^x + 2^{-x} $

Question

How many real solutions are there of the equation $2 \cos (\frac{x^2 + x}{6}) = 2^x + 2^{-x} $? Please illustrate.

Answer 1

AM-GM inequality says: $2^x + 2^{-x} \geq 2 \geq 2cos\left(\dfrac{x^2 + x}{6}\right)$. so equation occurs when: $2^x = 2^{-x}$ or $x = 0$, and this is the solution of the equation.

Answer 2

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 $\large x_2=-(1+6i\ln2)$. The real solution is only $\large\color{blue}{x=0}$.