How to find number of real roots of a transcendental equation? The number of real roots of the equation 
$$2\cos\left(\frac{x^2+x}6\right)=2^x+2^{-x}$$
Another question is...
can we use descartes rule of sign in here or in any transcendental equation ? 

Notice that
$$2\cos\left(\frac{x^2+x}6\right) \le 2$$
$$2^x+2^{-x} \ge 2$$
First is true since $\cos\left(\frac{x^2+x}6\right) \le 1$ and the second is equivalent to $\left(2^{\frac{x}{2}} - \frac{1}{2^{\frac{x}{2}}} \right)^2 \ge 0$ which is always true. So in order to have an equality the second inequality also has to be an equality and happens only when $2^{\frac{x}{2}} = \frac{1}{2^{\frac{x}{2}}} \leftrightarrow2^x=1\leftrightarrow x = 0$.
