Let y = y(x) be the solution of the differential equation $$\sec x \frac{dy}{dx}-2y=2+3\sin x, x\epsilon \left(-\frac{\pi}{2}, \frac{\pi}{2} \right), y(0)=-\frac{7}{4}$$. Then $$y\left(\frac{\pi}{6}\right)$$ is equal to:

The ODE is $$\sec x \frac{dy}{dx} - 2y = 2 + 3\sin x$$, with $$y(0) = -7/4$$.

Rewrite: $$\frac{dy}{dx} - 2y\cos x = (2 + 3\sin x)\cos x = 2\cos x + 3\sin x\cos x$$.

Integrating factor = $$e^{-2\int \cos x \, dx} = e^{-2\sin x}$$.

Solution: $$ye^{-2\sin x} = \int (2\cos x + 3\sin x \cos x)e^{-2\sin x}dx$$

Let $$t = \sin x$$, $$dt = \cos x \, dx$$:

$$= \int (2 + 3t)e^{-2t}dt = 2\int e^{-2t}dt + 3\int te^{-2t}dt$$

$$\int e^{-2t}dt = -\frac{e^{-2t}}{2}$$

$$\int te^{-2t}dt = -\frac{te^{-2t}}{2} + \frac{1}{2}\int e^{-2t}dt = -\frac{te^{-2t}}{2} - \frac{e^{-2t}}{4}$$

So: $$\int (2+3t)e^{-2t}dt = -e^{-2t} - \frac{3te^{-2t}}{2} - \frac{3e^{-2t}}{4} = e^{-2t}\left(-1 - \frac{3t}{2} - \frac{3}{4}\right) = e^{-2t}\left(-\frac{7}{4} - \frac{3t}{2}\right)$$

$$ye^{-2\sin x} = e^{-2\sin x}\left(-\frac{7}{4} - \frac{3\sin x}{2}\right) + C$$

Apply $$y(0) = -7/4$$:

$$-\frac{7}{4}e^0 = e^0\left(-\frac{7}{4}\right) + C \Rightarrow C = 0$$

So $$y = -\frac{7}{4} - \frac{3\sin x}{2}$$.

$$y(\pi/6) = -\frac{7}{4} - \frac{3}{2} \cdot \frac{1}{2} = -\frac{7}{4} - \frac{3}{4} = -\frac{10}{4} = -\frac{5}{2}$$

The answer is Option 4: $$-\frac{5}{2}$$.