Answer: (a) \(x=\pi\). (b) \(x=2.30\) and \(x=3.98\).
Answer: (a) \(x=\pi\). (b) \(x=2.30\) and \(x=3.98\).
(a) Differentiate
\(f(x)=3\sin^{2}x-2\cos x\).
This gives
\(f'(x)=6\sin x\cos x+2\sin x\).
Factorising,
\(f'(x)=2\sin x(3\cos x+1)\).
At a stationary point, \(f'(x)=0\). Thus
\(2\sin x(3\cos x+1)=0\).
So \(\sin x=0\) or \(\cos x=-\frac13\).
In the interval \(2\le x\le4\), \(\sin x=0\) gives
\(x=\pi\).
The solutions of \(\cos x=-\frac13\) are outside the interval \(2\le x\le4\). Therefore the stationary point has
\(x=\pi\).
(b) Substitute the definition of \(f(x)\):
\(3\sin^{2}x-2\cos x=1-3\cos x\).
Using \(\sin^{2}x=1-\cos^{2}x\),
\(3(1-\cos^{2}x)-2\cos x=1-3\cos x\).
Simplifying gives
\(3\cos^{2}x-\cos x-2=0\).
Factorising,
\((3\cos x+2)(\cos x-1)=0\).
So \(\cos x=-\frac23\) or \(\cos x=1\).
There are no solutions from \(\cos x=1\) in \(2\le x\le4\). For \(\cos x=-\frac23\), the solutions in the interval are
\(x=2.30\) and \(x=3.98\) to 3 significant figures.
