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Math Help - problem on equation

  1. #1
    Senior Member Sambit's Avatar
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    Question 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
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  2. #2
    MHF Contributor
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    Hint: 2^x + 2^{-x} has the global minimum at x = 0.

    See also this equation at WolframAlpha.
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