If y = y x is the solution of the equation e sin y cos y d y d x + e sin y cos x = cos x , y 0 = 0 ; then 1…

$\Rightarrow e^{\sin y}\cos y\frac{dy}{dx}=\frac{dt}{dx}$

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

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

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

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

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

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

$\Rightarrow y=0$

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