${e^{({{\cos }^2}x + {{\cos }^4}x + ...........\infty )\ln 2}} = {2^{{{\cos }^2}x + {{\cos }^4}x + ...........\infty }}$ = ${2^{{{{{\cos }^2}x} \over {1 - {{\cos }^2}x}}}}$ $= {2^{{{\cot }^2}x}}$ Given, ${t^2} - 9t + 8 = 0 \Rightarrow t = 1,8$ $\Rightarrow {2^{{{\cot }^2}x}} = 1,8 \Rightarrow co{t^2}x = 0,3$ $0 < x < {\pi \over 2} \Rightarrow \cot x = \sqrt 3$ $\therefore$ ${{2\sin x} \over {\sin x + \sqrt 3 \cos x}} = {2 \over {1 + \sqrt 3 \cot x}} = {2 \over 4} = {1 \over 2}$