If $$y = y(x)$$ is the solution of the equation $$e^{\sin y}\cos y \frac{dy}{dx} + e^{\sin y}\cos x = \cos x$$, $$y(0) = 0$$; then $$1 + y\left(\frac{\pi}{6}\right) + \frac{\sqrt{3}}{2}y\left(\frac{\pi}{3}\right) + \frac{1}{\sqrt{2}}y\left(\frac{\pi}{4}\right)$$ is equal to ______.

The differential equation is $$e^{\sin y}\cos y \frac{dy}{dx} + e^{\sin y}\cos x = \cos x$$.

Let $$t = e^{\sin y}$$. Then $$\frac{dt}{dx} = e^{\sin y}\cos y \frac{dy}{dx}$$.

The equation becomes $$\frac{dt}{dx} + t\cos x = \cos x$$, which is a linear first-order ODE.

The integrating factor is $$e^{\int \cos x \, dx} = e^{\sin x}$$.

Multiplying both sides by the integrating factor: $$\frac{d}{dx}(t \cdot e^{\sin x}) = \cos x \cdot e^{\sin x}$$.

Integrating: $$t \cdot e^{\sin x} = \int \cos x \cdot e^{\sin x} \, dx = e^{\sin x} + C$$.

So $$t = 1 + Ce^{-\sin x}$$, i.e., $$e^{\sin y} = 1 + Ce^{-\sin x}$$.

Using the initial condition $$y(0) = 0$$: $$e^{\sin 0} = 1 + Ce^{-\sin 0}$$, so $$1 = 1 + C$$, giving $$C = 0$$.

Therefore $$e^{\sin y} = 1$$, which means $$\sin y = 0$$, so $$y = n\pi$$ for integer $$n$$. By continuity and $$y(0) = 0$$, we have $$y(x) = 0$$ for all $$x$$.

Thus $$y\left(\frac{\pi}{6}\right) = 0$$, $$y\left(\frac{\pi}{3}\right) = 0$$, and $$y\left(\frac{\pi}{4}\right) = 0$$.

The expression evaluates to $$1 + 0 + \frac{\sqrt{3}}{2} \cdot 0 + \frac{1}{\sqrt{2}} \cdot 0 = 1$$.