What is the number of real roots of the equation 2\cos(\frac{x^2+x}{6})=2^x+2^{-x}

Beverley Rahman

Beverley Rahman

Answered question

2022-02-26

What is the number of real roots of the equation
2cos(x2+x6)=2x+2x

Answer & Explanation

Anderson Higgs

Anderson Higgs

Beginner2022-02-27Added 5 answers

So:
2x+2x22x2x=1
2x+2x2
with equality only when 2x=2x, i.e. x=0
Also,
2cos(x2+x6)2
So the only solution is when LHS=RHS=2. The only olution for RHS=2 is x=0, which also turns out to be a solution for LHS.
Hence there is exactly one real root, x=0.

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