Answer: \(k=\pi-1\).

Answer: \(k=\pi-1\).

The curve is \(y=1+\cos x\). One complete period runs from \(x=0\) to \(x=2\pi\).

The line \(y=1\) intersects the curve when

\(1+\cos x=1\),

so \(\cos x=0\). In one period, this occurs at

\(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\).

Area \(A\) is between the line \(y=1\) and the curve where the curve lies below the line, so

Thus

Therefore

\(A=[-\sin x]_{\pi/2}^{3\pi/2}=1-(-1)=2\).

The tangent line \(PQ\) is horizontal at the maximum value of the curve, so \(PQ\) is the line \(y=2\).

The rectangle between \(y=1\) and \(y=2\), from \(x=0\) to \(x=2\pi\), has area

\(2\pi\).

The parts between the curve and \(y=1\) where the curve is above \(y=1\) have total area

This is

\((1-0)+(0-(-1))=2\).

Hence

\(B=2\pi-2\).

Since

\(A:B=1:k\),

we have

\(2:(2\pi-2)=1:k\).

Therefore

\(\displaystyle k=\frac{2\pi-2}{2}=\pi-1\).