How to solve this equation $2\cos(\frac {x^2+x}{6})=2^x+2^{-x}$ How do I solve for $x$ from this equation?
$$2\cos\frac {x^2+x}{6}=2^x+2^{-x}$$
 A: $-2\le2\cos(\frac{x^2 + x}{6})\le 2$ and $2\le2^x + 2^{-x}$.
Therefore the only possible solution is when both equal $2$.
A: Get bounds for the left hand side and the right hand side separately ...
A: 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 $\large x_2=-(1+6i\ln2)$.
A: 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 ${2^x + 2^{-x} > 2}$. :))
