Solve the equations :xcos3y+ 3xcosysin2y= 14xsin3y+ 3xcos2ysiny= 13

Solution

The given equations are

xcos3y + 3x cosy sin2y = 14 …(i)

and xsin3y + 3x cos2y siny = 13 …(ii)

Adding (i) and (ii), we have

x(cos3y + 3cosy sin2y + 3cos2y siny + sin3y) = 27

or x1/3 (cosy + siny) = 3 …(iii)

Subtracting (ii) from (i), we have

x(cos3y + 3cosy sin2y − 3cos2y siny − sin3y) = 1

or x(cosy − sin)3 = 1

or x1/3 (cosy − siny) = 1 …(iv)

Now adding and subtracting (iii) and (iv), we have

2x1/3 cosy = 4 and 2x1/3 siny = 2

or x1/3 cosy = 2 …(v)

and x1/3 siny = 1 …(vi)

Dividing (vi) by (v), we have tany = 1/2

∴ either (i) siny = and cosy =

or (ii) siny = and cosy = −

In case (i), y = 2nπ + α

where 0 < α < π/2 and sinα =

i.e. y lies in the first quadrant.

∴ from (vi), x1/3 . () = 1  or x =

In case (ii), y = 2nπ + (π + α),

where 0 < α < π/2 and sinα =

i.e. y lies in the 3rd quadrant .

∴ from (vi), x1/3 (−) = 1   or x = .

Hence the required solution is

, y = 2nπ + α,     ∀ n ∈ I

or