(a) Using double-angle identities, \[f(\theta)=6\sin2\theta+8\cos2\theta+8.\] Comparing with \[R\cos(2\theta-\alpha)=R\cos\alpha\cos2\theta+R\sin\alpha\sin2\theta\] gives \[R=\sqrt{8^2+6^2}=10,\quad k=8,\quad\tan\alpha=\frac68=\frac34.\] Thus \(\alpha=\tan^{-1}(3/4)\approx0.6435\) radians, and \[f(\theta)=10\cos(2\theta-\alpha)+8.\]

(b) \[\cos(2\theta-\alpha)=\frac{17-8}{10}=0.9.\] Let \(\beta=\cos^{-1}(0.9)\). The two first positive roots are \((\alpha-\beta)/2\) and \((\alpha+\beta)/2\); the smaller is \[\theta=\frac{\tan^{-1}(3/4)-\cos^{-1}(0.9)}2\approx0.0962\text{ radians}.\]

(c) \[\int f(\theta)\,d\theta=-3\cos2\theta+4\sin2\theta+8\theta+C.\] Equivalently, \(5\sin(2\theta-\alpha)+8\theta+C\), with exact \(\alpha=\tan^{-1}(3/4)\).