The number of real roots of the equation
$$2cos \left( \frac {x^2 + x} {6} \right)=2^x + 2^{-x}$$
Answer options are : 0,1,2,∞
My approach :
range of cos function is [-1,1]
thus the RHS of the equation belongs to [-2,2]
So, we have
-2 ≤ 2x + 2-x ≤ 2
solving the right inequality, i got 2x...