If y=y(x) is the solution of the equation e^{ | vedclass

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(D) Let $e^{\sin y} = t$. Then,differentiating with respect to $x$,we get $e^{\sin y} \cos y \frac{dy}{dx} = \frac{dt}{dx}$. The given differential eq

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(D) Let $e^{\sin y} = t$. Then,differentiating with respect to $x$,we get $e^{\sin y} \cos y \frac{dy}{dx} = \frac{dt}{dx}$. The given differential equation becomes $\frac{dt}{dx} + t \cos x = \cos x$. This is a linear differential equation of the form $\frac{dt}{dx} + P(x)t = Q(x)$,where $P(x) = \cos x$ and $Q(x) = \cos x$. The integrating factor $(I.F.)$ is $e^{\int \cos x \, dx} = e^{\sin x}$. The solution is $t \cdot e^{\sin x} = \int \cos x \cdot e^{\sin x} \, dx$. Let $u = \sin x$,then $du = \cos x \, dx$. So,$t \cdot e^{\sin x} = \int e^u \, du = e^u + c = e^{\sin x} + c$. Substituting $t = e^{\sin y}$,we get $e^{\sin y} \cdot e^{\sin x} = e^{\sin x} + c$. Given $y(0) = 0$,we have $e^{\sin 0} \cdot e^{\sin 0} = e^{\sin 0} + c$,which implies $1 \cdot 1 = 1 + c$,so $c = 0$. Thus,$e^{\sin y} \cdot e^{\sin x} = e^{\sin x}$,which simplifies to $e^{\sin y} = 1$. Taking the natural logarithm on both sides,$\sin y = 0$,which implies $y = 0$ for all $x$. Therefore,$y\left(\frac{\pi}{6}\right) = 0$,$y\left(\frac{\pi}{3}\right) = 0$,and $y\left(\frac{\pi}{4}\right) = 0$. The expression $1 + y\left(\frac{\pi}{6}\right) + \frac{\sqrt{3}}{2} y\left(\frac{\pi}{3}\right) + \frac{1}{\sqrt{2}} y\left(\frac{\pi}{4}\right) = 1 + 0 + 0 + 0 = 1$.

If $y=y(x)$ is the solution of the equation $e^{\sin y} \cos y \frac{dy}{dx} + e^{\sin y} \cos x = \cos x$ with $y(0)=0$,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

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