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The number of the real roots of the equation $$2 \cos (x(x + 1)) = 2^x + 2^{-x}$$ is
$$2 \cos (x(x + 1)) = 2^x + 2^{-x}$$
The maximum value of LHS is 2 when $$\cos (x(x + 1))$$ is 1 and the minimum value of RHS is 2 using AM $$\geq$$ GM
Hence LHS and RHS can only be equal when both sides are 2. For LHS, cosx(x+1)=1 => x(x+1)=0 => x=0,-1
For RHS minimum value, x=0
Hence only one solution x=0
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