The correct option is D 6
xcos3y+3xcosysin2y=14 ⋯(1)
xsin3y+3xcos2ysiny=13 ⋯(2)
Adding Eqs. (1) and (2)
x(cos3y+3cos2ysiny+3cosysin2y+sin3y)=27
⇒x(cosy+siny)3=27
⇒x1/3(cosy+siny)=3 ⋯(3)
Subtracting Eq (2) from Eq (1), we get
x(cos3y+3cosysin2y−3cos2ysiny−sin3y)=1
⇒x(cosy−siny)3=1
⇒x1/3(cosy−siny)=1 ⋯(4)
Dividing Eq (3) by Eq (4), we get
cosy+siny=3cosy−3siny
⇒tany=12
tan function is positive in 1st quadrant and 3rd quadrant.
Therefore, there are exactly six values of y∈[0,6π]