(a) Expanding gives \[3\sqrt3\sin(\theta+\pi/6)-2\sin\theta=\frac52\sin\theta+\frac{3\sqrt3}2\cos\theta.\] Compare with \(R\cos\alpha\sin\theta+R\sin\alpha\cos\theta\): \[R=\sqrt{\frac{25+27}4}=\sqrt{13},\qquad\alpha=\tan^{-1}\left(\frac{3\sqrt3}{5}\right)=0.805.\]

(b) Use unrounded \(\alpha\). Then \[\sin(2x+\alpha)=\sqrt{\frac6{13}}.\] Put \(\gamma=\sin^{-1}\sqrt{6/13}=0.746898\ldots\). In \(\alpha<2x+\alpha<2\pi+\alpha\), the two valid angles are \(\pi-\gamma\) and \(2\pi+\gamma\); \(\gamma<\alpha\) would give \(x<0\). Thus \[x=\frac{\pi-\gamma-\alpha}{2}=0.795,\qquad x=\frac{2\pi+\gamma-\alpha}{2}=3.11.\]