What can be said about the number of roots of the equation, given by:- $\displaystyle 2 cos(\frac{x^2+x}{6}) = 2^x + 2^{-x} ? $
there are four options:-
(a) no root
(b) 1 root
(c) 2 roots
(d) infinitely many roots
Hint: $\displaystyle 2^x + 2^{-x}$ has the global minimum at x = 0.
See also this equation at WolframAlpha.