If y = y(x) is the solution
Question:

If $y=y(x)$ is the solution of the equaiton $e^{\sin y} \cos y \frac{d y}{d x}+e^{\sin y} \cos x=\cos x, y(0)=0$;

then $1+\mathrm{y}\left(\frac{\pi}{6}\right)+\frac{\sqrt{3}}{2} \mathrm{y}\left(\frac{\pi}{3}\right)+\frac{1}{\sqrt{2}} \mathrm{y}\left(\frac{\pi}{4}\right) \quad$ is equal to

Solution:

Put $\mathrm{e}^{\text {siny }}=\mathrm{t}$

$\Rightarrow e^{\sin y} \cos y \frac{d y}{d x}=\frac{d t}{d x}$

$\Rightarrow$ D.E is $\frac{\mathrm{dt}}{\mathrm{dx}}+\mathrm{t} \cos \mathrm{x}=\cos \mathrm{x}$

I.F. $=e^{\int \cos x d x}=e^{\sin x}$

$\Rightarrow$ solution is t.e $^{\sin x}=\int \cos x e^{\sin x}$

$\Rightarrow$ solution is t.e $^{\sin x}=\int \cos x e^{\sin x}$

$\Rightarrow e^{\sin y} e^{\sin x}=e^{\sin x}+c$

$\because x=0, y=0 \Rightarrow c=0$

$\Rightarrow e^{\sin y}=1$

$\Rightarrow y=0$

$\Rightarrow 1+y\left(\frac{\pi}{6}\right)+\frac{\sqrt{3}}{2} y\left(\frac{\pi}{3}\right)+\frac{1}{\sqrt{2}} y\left(\frac{\pi}{4}\right)=1$

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