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

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- Oct 19th 2010, 07:51 AMSambitproblem on equation
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 - Oct 19th 2010, 08:20 AMemakarov
Hint: $\displaystyle 2^x + 2^{-x}$ has the global minimum at x = 0.

See also this equation at WolframAlpha.