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(D) The given differential equation is $\cos x(\ln(\cos x))^2 dy + (\sin x - 3y \sin x \ln(\cos x)) dx = 0$. Dividing by $dx \cdot \cos x(\ln(\cos x))

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$\frac{1}{\ln 3 - \ln 4}$

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(D) The given differential equation is $\cos x(\ln(\cos x))^2 dy + (\sin x - 3y \sin x \ln(\cos x)) dx = 0$. Dividing by $dx \cdot \cos x(\ln(\cos x))^2$,we get: $\frac{dy}{dx} - \frac{3 \sin x \ln(\cos x)}{\cos x(\ln(\cos x))^2} y = -\frac{\sin x}{\cos x(\ln(\cos x))^2}$ $\frac{dy}{dx} - \frac{3 	an x}{\ln(\cos x)} y = -\frac{	an x}{(\ln(\cos x))^2}$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{3 	an x}{\ln(\cos x)}$. Integrating factor $I.F. = e^{\int P(x) dx} = e^{\int -\frac{3 	an x}{\ln(\cos x)} dx}$. Let $u = \ln(\cos x)$,then $du = \frac{1}{\cos x} (-\sin x) dx = -	an x dx$. So,$I.F. = e^{3 \int \frac{1}{u} du} = e^{3 \ln u} = u^3 = (\ln(\cos x))^3$. The solution is $y \cdot I.F. = \int Q(x) \cdot I.F. dx + C$. $y(\ln(\cos x))^3 = \int -\frac{	an x}{(\ln(\cos x))^2} \cdot (\ln(\cos x))^3 dx + C$. $y(\ln(\cos x))^3 = -\int 	an x \ln(\cos x) dx + C$. Let $v = \ln(\cos x)$,then $dv = -	an x dx$. $y(\ln(\cos x))^3 = \int v dv + C = \frac{v^2}{2} + C = \frac{(\ln(\cos x))^2}{2} + C$. Given $y(\frac{\pi}{4}) = -\frac{1}{\ln 2}$. Since $\ln(\cos \frac{\pi}{4}) = \ln(\frac{1}{\sqrt{2}}) = -\frac{1}{2} \ln 2$. $(-\frac{1}{\ln 2})(-\frac{1}{2} \ln 2)^3 = \frac{1}{2} (-\frac{1}{2} \ln 2)^2 + C$. $(-\frac{1}{\ln 2})(-\frac{1}{8} (\ln 2)^3) = \frac{1}{2} (\frac{1}{4} (\ln 2)^2) + C$. $\frac{1}{8} (\ln 2)^2 = \frac{1}{8} (\ln 2)^2 + C \implies C = 0$. Thus,$y(\ln(\cos x))^3 = \frac{(\ln(\cos x))^2}{2} \implies y = \frac{1}{2 \ln(\cos x)}$. For $x = \frac{\pi}{6}$,$y(\frac{\pi}{6}) = \frac{1}{2 \ln(\cos \frac{\pi}{6})} = \frac{1}{2 \ln(\frac{\sqrt{3}}{2})} = \frac{1}{2(\frac{1}{2} \ln 3 - \ln 2)} = \frac{1}{\ln 3 - \ln 4}$.

Let $y=y(x)$ be the solution of the differential equation $\cos x(\ln(\cos x))^2 dy + (\sin x - 3y \sin x \ln(\cos x)) dx = 0$,where $x \in (0, \frac{\pi}{2})$. If $y(\frac{\pi}{4}) = \frac{-1}{\ln 2}$,then $y(\frac{\pi}{6})$ is:

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