If $$\cos x \frac{dy}{dx} - y\sin x = 6x$$, $$(0 < x < \frac{\pi}{2})$$ and $$y\left(\frac{\pi}{3}\right) = 0$$, then $$y\left(\frac{\pi}{6}\right)$$ is equal to:

We have the differential equation

$$\cos x \dfrac{dy}{dx} - y \sin x = 6x, \qquad 0 < x < \dfrac{\pi}{2}.$$

First we rewrite it in the standard linear form. Dividing every term by $$\cos x$$ gives

$$\dfrac{dy}{dx} - y \tan x = 6x \sec x.$$

A first-order linear equation of the form $$\dfrac{dy}{dx} + P(x)\,y = Q(x)$$ is solved by multiplying with the integrating factor $$\mu(x)=e^{\int P(x)\,dx}.$$

Here $$P(x) = -\tan x,$$ so

$$\mu(x) = e^{\displaystyle\int -\tan x\,dx} = e^{-\ln|\sec x|} = e^{\ln|\cos x|} = \cos x.$$

Multiplying the whole differential equation by $$\cos x$$ (the integrating factor) we obtain

$$\cos x \dfrac{dy}{dx} - y \sin x = 6x,$$

which is exactly the left-hand side of our original equation, confirming that $$\cos x$$ is indeed the correct integrating factor.

By construction, the left side is now the derivative of the product $$y\cos x$$, because

$$\dfrac{d}{dx}\!\bigl(y\cos x\bigr)=\cos x\dfrac{dy}{dx}-y\sin x.$$

So we can rewrite the equation as

$$\dfrac{d}{dx}\!\bigl(y\cos x\bigr)=6x.$$

Integrating both sides with respect to $$x$$, we get

$$y\cos x=\int 6x\,dx = 3x^{2}+C,$$

where $$C$$ is the constant of integration.

We are given the initial condition $$y\!\left(\dfrac{\pi}{3}\right)=0.$$ Substituting $$x=\dfrac{\pi}{3}$$ and $$y=0$$ in the integrated result:

$$0\cdot\cos\!\left(\dfrac{\pi}{3}\right)=3\!\left(\dfrac{\pi}{3}\right)^{2}+C,$$

$$0=\dfrac{3\pi^{2}}{9}+C=\dfrac{\pi^{2}}{3}+C.$$

Hence

$$C=-\dfrac{\pi^{2}}{3}.$$

Substituting $$C$$ back, the particular solution becomes

$$y\cos x = 3x^{2}-\dfrac{\pi^{2}}{3},$$

so

$$y(x)=\dfrac{3x^{2}-\dfrac{\pi^{2}}{3}}{\cos x}.$$

Now we must find $$y\!\left(\dfrac{\pi}{6}\right).$$ Putting $$x=\dfrac{\pi}{6}$$:

Numerator:

$$3\!\left(\dfrac{\pi}{6}\right)^{2}-\dfrac{\pi^{2}}{3}=3\cdot\dfrac{\pi^{2}}{36}-\dfrac{\pi^{2}}{3} =\dfrac{\pi^{2}}{12}-\dfrac{\pi^{2}}{3} =\pi^{2}\!\left(\dfrac{1}{12}-\dfrac{4}{12}\right) =-\dfrac{3\pi^{2}}{12} =-\dfrac{\pi^{2}}{4}.$$

Denominator:

$$\cos\!\left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}.$$

Therefore

$$y\!\left(\dfrac{\pi}{6}\right)=\dfrac{-\dfrac{\pi^{2}}{4}}{\dfrac{\sqrt{3}}{2}} =-\dfrac{\pi^{2}}{4}\cdot\dfrac{2}{\sqrt{3}} =-\dfrac{\pi^{2}}{2\sqrt{3}}.$$

Hence, the correct answer is Option D.