Answer: (a) \(x=-0.257\) or \(x=0.304\). (b) \(x=0.0236\) or \(x=0.469\).

Let

\(u=3x+1.5.\)

Since \(-0.3\leqslant x\leqslant0.5\),

\(0.6\leqslant u\leqslant3.0.\)

(a) The equation \(\mathrm{f}(x)=0\) is

\(15\cos^2u+7\sin u-13=0.\)

Use \(\cos^2u=1-\sin^2u\):

\(15(1-\sin^2u)+7\sin u-13=0.\)

So

\(15\sin^2u-7\sin u-2=0.\)

Factorising,

\((5\sin u+1)(3\sin u-2)=0.\)

Thus \(\sin u=-\frac15\) or \(\sin u=\frac23\).

In the interval \(0.6\leqslant u\leqslant3.0\), only \(\sin u=\frac23\) gives solutions. Hence

\(u=\sin^{-1}\frac23\quad\text{or}\quad u=\pi-\sin^{-1}\frac23.\)

Since \(x=\frac{u-1.5}{3}\),

\(x=-0.2567\ldots\quad\text{or}\quad x=0.3039\ldots.\)

So

\(x=-0.257\quad\text{or}\quad x=0.304.\)

(b) Differentiate:

\(\mathrm{f}'(x)=15\cdot2\cos u(-3\sin u)+7(3\cos u).\)

So

\(\mathrm{f}'(x)=3\cos u(7-30\sin u).\)

At a stationary point,

\(3\cos u(7-30\sin u)=0.\)

Thus \(\cos u=0\) or \(\sin u=\frac7{30}\).

In the interval \(0.6\leqslant u\leqslant3.0\), these give

\(u=\frac{\pi}{2}\quad\text{or}\quad u=\pi-\sin^{-1}\frac7{30}.\)

Therefore

\(x=\frac{\frac{\pi}{2}-1.5}{3}=0.0236\ldots,\)

or

\(x=\frac{\pi-\sin^{-1}\frac7{30}-1.5}{3}=0.4686\ldots.\)

So the two stationary point \(x\)-coordinates are

\(x=0.0236\quad\text{and}\quad x=0.469.\)