If $\cos 3\mathrm{\theta }=\cos 3\mathrm{\alpha }$, then the value of sin $\mathrm{\theta }$ can be given by

$\cos 3\mathrm{\theta }=\cos 3\mathrm{\alpha }$
Putting n = 0, 1, we have
$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\mathrm{\alpha }=2\mathrm{n\pi }\pm 3\mathrm{\alpha }\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\mathrm{\theta }=3\mathrm{\alpha }\text{ or }-3\mathrm{\alpha }\text{ or }2\mathrm{\pi }+3\mathrm{\alpha }\text{ or }2\mathrm{\pi }-3\mathrm{\alpha }\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\theta }=\mathrm{\alpha }\text{ or }-\mathrm{\alpha }\text{ or }\frac{2\mathrm{\pi }}{3}+\mathrm{\alpha }\text{ or }\frac{2\mathrm{\pi }}{3}-\mathrm{\alpha }\end{array}$
Hence, options (a), (c), (d) are correct
If $\mathrm{n}=-1,\text{ then }3\mathrm{\theta }=-2\mathrm{\pi }\pm 3\mathrm{\alpha }$
$$\begin{gathered} \Rightarrow \mathrm{\theta }=-\frac{2\mathrm{\pi }}{3}\pm \mathrm{\alpha } \\ \begin{array}{r}\Rightarrow \sin \mathrm{\theta }=\sin \left(-\frac{2\mathrm{\pi }}{3}\pm \mathrm{\alpha }\right)=-\sin \left(\frac{2\mathrm{\pi }}{3}\pm \mathrm{\alpha }\right)\\ =-\sin \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\pm \mathrm{\alpha }\right)=-\sin \left(\frac{\mathrm{\pi }}{3}\pm \mathrm{\alpha }\right)\end{array} \end{gathered}$$
Hence, (b) is not correct.