# problem on equation

• Oct 19th 2010, 07:51 AM
Sambit
problem on equation
What can be said about the number of roots of the equation, given by:- $2 cos(\frac{x^2+x}{6}) = 2^x + 2^{-x} ?$

there are four options:-
(a) no root
(b) 1 root
(c) 2 roots
(d) infinitely many roots
• Oct 19th 2010, 08:20 AM
emakarov
Hint: $2^x + 2^{-x}$ has the global minimum at x = 0.