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Let \(f(x) = 6 + 6 \sin x\).

Part of the graph of \(f\) is given below.

The figure is not to scale.

The shaded region is bounded by the curve of \(f\), the x-axis, and the y-axis.

1. Solve, for \(0 \leq x \leq 2\pi\)

   1. \begin{equation*} 6 + 6 \sin x = 6 \end{equation*}
   2. \begin{equation*} 6 + 6 \sin x = 0 \end{equation*}

2. Write down the exact value of the x-intercept of \(f\), for \(0 \leq x \leq 2\pi\).

3. The area of the shaded region is \(k\). Find the value of \(k\), giving your answer in terms of \(\pi\).

Let \(g(x) = 6 + 6 \sin (x - \frac{\pi}{2})\). The graph of \(f\) is transformed into that of \(g\).

4. Give a full geometric description of this transformation.
5. Given that \(\int_p^{p+\frac{3\pi}{2}}g(x)\,dx = k\) and \(0 \leq p < 2\pi\), find the two values of \(p\).