Prove that
\[\begin{equation*}
\frac{d}{dx} (3 \cos x - \cos 3x)= 6 \sin x \cos 2x.
\end{equation*}\]

Using the chain rule or by thinking about horizontal scaling, we have that
\[\begin{equation*}
\frac{d}{dx} \cos 3x = - 3 \sin 3x.
\end{equation*}\]
By the compound angle formula,
\[\begin{align*}
\sin 3x = \sin(x + 2x) &= \sin x \cos 2x + \sin 2x \cos x \\
&= \sin x \cos 2x + 2 \sin x \cos^2 x \\
&= \sin x \cos 2x + \sin x (1 + \cos 2x) \\
&= 2 \sin x \cos 2x + \sin x
\end{align*}\]
so that
\[\begin{align*}
\frac{d}{dx} \left( 3 \cos x - \cos 3x \right) &= - 3 \sin x + 3 \sin 3x \\
&= - 3 \sin x + 6 \sin x \cos 2x + 3 \sin x \\
&= 6 \sin x \cos 2x.
\end{align*}\]

Alternatively, we can use the standard formula, \(\sin P-\sin Q = 2\cos\left(\dfrac{P+Q}{2}\right)\sin\left(\dfrac{P-Q}{2}\right)\).

For what two ranges of values of \(x\) between \(0\) and \(\pi\) is
\[\begin{equation*}
3 \cos x - \cos 3x
\end{equation*}\]

an increasing function of \(x\)?

A function is increasing exactly when its derivative is positive, so we are looking for those ranges of values of \(x\) such that
\[\begin{equation*}
\frac{d}{dx} \left( 3 \cos x - \cos 3x \right) > 0
\end{equation*}\]
which, from the above, is equivalent to the inequality
\[\begin{equation*}
\sin x \cos 2x > 0.
\end{equation*}\]

This holds exactly when \(\sin x\) and \(\cos 2x\) have the same sign. We know that \(\sin x\) is positive when \(0 < x < \pi\), so the required inequality can only hold when \(\cos 2x > 0\). We know that:

\(\cos 2x > 0\) when \(0 < x < \pi/4\),

\(\cos 2x < 0\) when \(\pi/4 < x < 3\pi/4\), and

\(\cos 2x > 0\) when \(3\pi/4 < x < \pi\).

The regions are therefore \(0< x < \pi/4\) and \(3\pi/4 < x < \pi\).

Draw a rough sketch-graph of \(y = 3 \cos x - \cos 3x\) for values of \(x\) from \(0\) to \(\pi\), indicating clearly any maximum or minimum points.

From the above, we know that \(y\) is

equal to \(2\) when \(x = 0\);

increasing in the region \(0 < x < \pi/4\);

stationary when \(x = \pi/4\);

decreasing in the region \(\pi/4 < x < 3\pi/4\);

stationary when \(x = 3\pi/4\);

increasing when \(3\pi/4 < x < \pi\); and

equal to \(-2\) when \(x = \pi\).

The only aspects left to consider are the location of the \(x\)-intercept in the decreasing phase and the value of the function at the stationary points. Symmetry suggests that the \(x\)-intercept might be at \(x=\frac{\pi}{2}\) so we start by trying that.
\[\begin{align*}
\text{At }x=\frac{\pi}{2},\quad y &= 3 \cos \frac{\pi}{2} - \cos \frac{3\pi}{2} = 0,\\
\text{At }x=\frac{\pi}{4},\quad y &= 3 \cos \frac{\pi}{4} - \cos \frac{3\pi}{4} = \frac{3\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 2\sqrt{2},\\
\text{At }x=\frac{3\pi}{4},\quad y &= 3 \cos \frac{3\pi}{4} - \cos \frac{9\pi}{4} = -\frac{3\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -2\sqrt{2}.
\end{align*}\]

We also have rotational symmetry about \(\left(\dfrac{\pi}{2},0\right)\), as \(y(x) = -y\left(\pi - x \right)\) for any \(0 < x < \dfrac{\pi}{2}\). We thus have the following sketch graph.