= Solution
Apply <Newton's method> to $F(t)=1-3\sin t/2-\sin^3t/3$. The derivative in part (a) gives
$$
\boxed{t_{k+1}=t_k+\frac{1-\frac32\sin t_k-\frac13\sin^3t_k}{\cos t_k\left(\frac32+\sin^2t_k\right)}.}
$$
Choose a starting point near a desired root, such as $t_0=0.6$ for the root in $(0,\pi/2)$; do not divide when $\cos t_k=0$. On $(0,\pi/2)$, $F$ is strictly decreasing and changes sign between its endpoints, so the <intermediate value theorem> gives exactly one root there. Its derivative is nonzero, giving the usual local convergence of <Newton's method>. The <periodic function> has other roots; the iteration does not impose global uniqueness or convergence from every starting value.
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