Resolvamos a equação quadrática $y^2 - 9y + 8 = 0$ : $\large{(y-\frac{9}{2})^2 = \frac{81}{4} - 8 = \frac{49}{4} = \left(\frac{7}{2} \right)^2}$ $\implies$ $\large{y = \frac{9\pm7}{2}} = \{1, \space 8\}$ . $\large{e^{(\sin^2 x + \sin^4 x + \sin^6 x + \cdots)\ln 2}}$ $=$ $\large{(e^{\ln 2})^{(\sin^2 x + \sin^4 x + \sin^6 x + \cdots)}}$ , logo: $\large{e^{(\sin^2 x + \sin^4 x + \sin^6 x + \cdots)\ln 2}}$ $=$ $\large{2^{(\sin^2 x + \sin^4 x + \sin^6 x + \cdots)}}$ Sabe-se que $\sin^2 x + \sin^4 x + \sin^6 x + \cdots = \large{\frac{\sin^2 x}{1-\sin^2 x}}$ . Assim: $\large{2^{\frac{\sin^2 x}{1-\sin^2 x}}}$ $=$ $1 = 2^0$ $\implies$ $\large{\frac{\sin^2 x}{1-\sin^2 x}} = 0$ $\implies$ $\sin x = 0$ $\color{red}{\text{(não \space convém)}}$ $\large{2^{\frac{\sin^2 x}{1-\sin^2 x}}}$ $=$ $8 = 2^3$ $\implies$ $\large{\frac{\sin^2 x}{1-\sin^2 x}} = 3$ $\implies$ $\sin^{2} x = 3 - 3\sin^{2} x$ , temos: $0 < \sin x = \large{\frac{\sqrt 3}{2}}$ $\implies$ $\cos x = \large{\frac{1}{2}}$ . O valor da razão $\Large{\frac{\cos x}{\cos x + \sin x}}$ $=$ $\Large{\frac{\frac{1}{2}}{\frac{1}{2} + \frac{\sqrt 3}{2}}}$ $=$ $\Large{\frac{1}{1 + \sqrt 3}}$ $=$ $\boxed{\large{\frac{\sqrt 3 - 1}{2}}}$