Answer on Trigonometry Question for sudhanshu
f'(x)=(cosx*2^sinx)/ln2-(sinx*2^cosx)/ln2=0. 2^sinx*cosx=2^cosx*sinx it's possible when cosx=sinx. As we're searching the minimum we interested in such
values of x when cosx and sinx are both negative. So x=5pi/4+kpi, where k is
random integer. The value of our expression in this case is 2^(1/sqrt(2)).
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