The equation of a normal to the curve siny = xsin(π/3 + y) at x = 0, is

The correct option is D
2x+√3y=0
siny = xsin(π/3 + y). (1)
Putting x = 0 ⇒ siny = 0 ⇒ y = 0
Differentiating (1) with respect to x,
cosy(dy/dx) = sin(π/3 + y) + xcos(π/3 + y)(dy/dx)
At x = 0, y = 0
cosy(dy/dx) = sin(π/3 + y) + xcos(π/3 + y)(dy/dx)
cos0(dy/dx) = sin(π/3) + 0
dy/dx = sin(π/3) = √3/2
Slope of the normal = -2/√3
Equation of the normal at (0, 0) is
y - 0 = (-2/√3)(x - 0)
√3y = -2x
2x + √3y = 0