Answer: \(k=-3\); normal \(y=-x+2\).

Use the quotient rule on

\(y=\frac{2+\sin 3x}{x+1}.\)

This gives

When \(x=\frac{\pi}{6}\),

and

Therefore

So

\(k=-3.\)

For the normal at \(x=0\), first find the point on the curve:

\(y=\frac{2+\sin 0}{1}=2.\)

So the point is \((0,2)\). The gradient of the curve at \(x=0\) is

\(\frac{(1)(3\cos0)-(2+\sin0)}{1^2}=3-2=1.\)

The normal gradient is the negative reciprocal, so it is \(-1\). Hence

\(y-2=-1(x-0),\)

and the normal is

\(y=-x+2.\)