esinycosydydx+esinycosx=cosx
Put esiny=t
⇒esiny×cosydydx=dtdx
Then, dtdx+tcosx=cosx
I.F. =e∫cosx dx=esinx
Solution of differential equation :
t⋅esinx=∫esinx⋅cosx dx
⇒esiny⋅esinx=esinx+C
At x=0,y=0
⇒1=1+C⇒C=0
∴siny+sinx=sinx
⇒y=0
⇒y(π6)=0, y(π3)=0, y(π4)=0
Required answer is 1+0+0+0=1