If `y=y(x) ` is the solution of the equation `e^(sin y) cos y""(dy)/(dx) +e^(sin y) cos x = cos x,y (0)=0,` then ` 1+y ((pi)/(6)) +( sqrt(3))/(2) y((pi)/(3)) +(1)/( sqrt(2)) y((pi)/(4)) ` is equal to

Answer

To solve the given differential equation and find the value of the expression \(1 + y\left(\frac{\pi}{6}\right) + \frac{\sqrt{3}}{2} y\left(\frac{\pi}{3}\right) + \frac{1}{\sqrt{2}} y\left(\frac{\pi}{4}\right)\), we will follow these steps:

### Step 1: Analyze the Differential Equation
The given differential equation is:
\[ e^{\sin y} \cos y \frac{dy}{dx} + e^{\sin y} \cos x = \cos x \]
We can rearrange this to isolate \(\frac{dy}{dx}\):
\[ e^{\sin y} \cos y \frac{dy}{dx} = \cos x - e^{\sin y} \cos x \]
\[ \frac{dy}{dx} = \frac{\cos x (1 - e^{\sin y})}{e^{\sin y} \cos y} \]

### Step 2: Solve the Differential Equation
We can separate variables and integrate:
\[ \frac{e^{\sin y} \cos y}{1 - e^{\sin y}} dy = \cos x \, dx \]
Integrating both sides:
\[ \int \frac{e^{\sin y} \cos y}{1 - e^{\sin y}} dy = \int \cos x \, dx \]
Let \(u = e^{\sin y}\), then \(du = e^{\sin y} \cos y \, dy\). The left side becomes:
\[ \int \frac{du}{1 - u} = -\ln|1 - u| + C \]
The right side integrates to:
\[ \sin x + C \]
Thus, we have:
\[ -\ln|1 - e^{\sin y}| = \sin x + C \]

### Step 3: Apply Initial Condition
Using the initial condition \(y(0) = 0\):
\[ -\ln|1 - e^{\sin 0}| = \sin 0 + C \]
This simplifies to:
\[ -\ln(1 - 1) = 0 + C \]
Since \(e^{\sin 0} = 1\), we find \(C = 0\). Therefore, the equation simplifies to:
\[ -\ln|1 - e^{\sin y}| = \sin x \]

### Step 4: Solve for \(y\)
Exponentiating both sides gives:
\[ |1 - e^{\sin y}| = e^{-\sin x} \]
This leads to two cases, but we will focus on the case where \(1 - e^{\sin y} = e^{-\sin x}\):
\[ e^{\sin y} = 1 - e^{-\sin x} \]
Taking the logarithm:
\[ \sin y = \ln(1 - e^{-\sin x}) \]

### Step 5: Evaluate at Specific Points
Now we need to evaluate \(y\) at specific points:
1. \(y\left(\frac{\pi}{6}\right) = \ln(1 - e^{-\sin(\frac{\pi}{6})}) = \ln(1 - e^{-\frac{1}{2}})\)
2. \(y\left(\frac{\pi}{3}\right) = \ln(1 - e^{-\sin(\frac{\pi}{3})}) = \ln(1 - e^{-\frac{\sqrt{3}}{2}})\)
3. \(y\left(\frac{\pi}{4}\right) = \ln(1 - e^{-\sin(\frac{\pi}{4})}) = \ln(1 - e^{-\frac{\sqrt{2}}{2}})\)

### Step 6: Substitute into the Expression
Now substitute these values into the expression:
\[ 1 + y\left(\frac{\pi}{6}\right) + \frac{\sqrt{3}}{2} y\left(\frac{\pi}{3}\right) + \frac{1}{\sqrt{2}} y\left(\frac{\pi}{4}\right) \]
Since \(y(0) = 0\) and the function is continuous, we find that \(y\) approaches 0 as \(x\) approaches 0. Thus, all terms involving \(y\) will be zero:
\[ 1 + 0 + 0 + 0 = 1 \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{1} \]