How to find number of real roots of a transcendental equation?

Question

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 ?

Answer 1

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 $2^{\frac{x}{2}} = \frac{1}{2^{\frac{x}{2}}} \leftrightarrow2^x=1\leftrightarrow x = 0$.

Answer 2

The right hand side,

$$2^x + 2^{-x}$$

is an even function that has a unique minimum value of $2$ in $x = 0$. The left hand side,

$$2\cos \left( \frac{x^2+x}{6}\right)$$

has $2$ as its maximal value. So there's only one candidate, and inspection shows that it is indeed a solution.

Answer 3

Note that $2^x+2^{-x}$ has to be in the interval $[-2,2]$ to cancel the LHS. But the image of $2^x+2^{-x}$ is $[2,\infty)$, so you only have to check the case when $2^x+2^{-x}=2$, and the only solution is when $x=0$.