find solution of th equation sin(2^x)cos(2^x) = (2^x + 2^(-x))/2

Question

to find the values of X

Answer 1

sin(2^x)cos(2^x)=(2^x+2^-x)/2,

2sin(2^x)cos(2^x)=2^x+2^-x, because 2sin(θ)cos(θ)=sin(2θ) for all θ:

sin(2^x+1)=2^x+2^-x. Sine must be in the range [-1,1]. When x=0, 2^x+2^-x=2, the minimum value, so there can be no solution.