C.

$\left\{\frac{2\mathrm{\pi }}{3}, \frac{4\mathrm{\pi }}{3}\right\}$

$$\begin{gathered} \mathrm{We} \mathrm{have}, \\ \left(5 + 4\cos \left(\mathrm{\theta }\right)\right)\left(2\cos \left(\mathrm{\theta }\right) + 1\right) = 0 ...\left(\mathrm{i}\right) \\ \cos \left(\mathrm{\theta }\right) = \frac{1 - {\tan }^2\left({\displaystyle \frac{\mathrm{\theta }}{2}}\right)}{1 + {\tan }^2\left(\frac{\mathrm{\theta }}{2}\right)}, \\ \therefore \cos \left(\mathrm{\theta }\right) = \frac{1 - {\mathrm{t}}^2}{1 + {\mathrm{t}}^2} \left[\because \mathrm{put} \tan \left(\frac{\mathrm{\theta }}{2}\right) = \mathrm{t}\right] \\ \mathrm{Then} \mathrm{Eq}.\left(\mathrm{i}\right) \mathrm{beecomes} \\ \left[5 + 4\left(\frac{1 - {\mathrm{t}}^2}{1 + {\mathrm{t}}^2}\right)\right]\left[2\left(\frac{1 - {\mathrm{t}}^2}{1 + {\mathrm{t}}^2}\right) +\hspace{0.17em}1\right] = 0 \\ \Rightarrow \left[5 + 5{\mathrm{t}}^2 + 4 - 4{\mathrm{t}}^2\right]\left[2 - 2{\mathrm{t}}^2 + 1 + {\mathrm{t}}^2\right] = 0 \\ \Rightarrow \left({\mathrm{t}}^2 + 9\right)\left(3 - {\mathrm{t}}^2\right) = 0 \\ \therefore \mathrm{t} = \pm \sqrt{3} \\ \Rightarrow \tan \left(\frac{\mathrm{\theta }}{2}\right) = \sqrt{3} \mathrm{or} \tan \left(\frac{\mathrm{\theta }}{2}\right) = - \sqrt{3} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{\mathrm{\theta }}{2} = \frac{\mathrm{\pi }}{3} or \frac{\mathrm{\theta }}{2} = \frac{2\mathrm{\pi }}{3} \\ \mathrm{\theta } = \frac{2\mathrm{\pi }}{3} \mathrm{or} \frac{4\mathrm{\pi }}{3} \end{gathered}$$

D.

No solution

$$\begin{gathered} \mathrm{We} \mathrm{have}, \\ \sqrt{3}\sin \left(\mathrm{x} \right)+ \cos \left(\mathrm{x} \right)= 4 \\ \Rightarrow \frac{\sqrt{3}}{2}\sin \left(\mathrm{x}\right)+\frac{1}{2}\cos \left(\mathrm{x}\right) = 2 \\ \Rightarrow \sin \left(\mathrm{x}\right)\cos \left(\frac{\mathrm{\pi }}{6}\right)+ \cos \left(\mathrm{x}\right)\sin \left(\frac{\mathrm{\pi }}{6}\right) = 2 \\ \Rightarrow \sin \left(\mathrm{x} + \frac{\mathrm{\pi }}{6}\right) = 2,\mathrm{which} \mathrm{is} \mathrm{not} \mathrm{possible} \\ \mathrm{Therefore},\mathrm{this} \mathrm{equation} \mathrm{has} \mathrm{no} \mathrm{solutions} \end{gathered}$$

B.

$\frac{2\mathrm{a}}{\mathrm{a} - 1}$

$$\begin{gathered} \mathrm{Given} \mathrm{that} \\ \frac{\tan \left(3\mathrm{A}\right)}{\tan \left(\mathrm{A}\right)} = \mathrm{a} \\ \Rightarrow \frac{3\tan \left(\mathrm{A}\right) - {\tan }^3\left(\mathrm{A}\right)}{\tan \left(\mathrm{A}\right)\left(1 - 3{\tan }^2\left(\mathrm{A}\right)\right)} = \mathrm{a} \\ \Rightarrow 3 - {\tan }^2\left(\mathrm{A}\right) = \mathrm{a} - 3{\tan }^2\left(\mathrm{A}\right) \\ \Rightarrow {\tan }^2\left(3\mathrm{a} - 1\right) = \mathrm{a} - 3 \\ \Rightarrow \tan \left(\mathrm{A}\right) = \pm \sqrt{\frac{\mathrm{a} - 3}{3\mathrm{a} - 1}} \\ \mathrm{Now}, \\ \frac{\sin \left(3\mathrm{A}\right)}{\sin \left(\mathrm{A}\right)} = \frac{3\sin \left(\mathrm{A}\right) - {\sin }^3\left(\mathrm{A}\right)}{\sin \left(\mathrm{A}\right)} \\ = 3 - 4{\sin }^2\left(\mathrm{A}\right) = 3 - 4\left(\frac{\mathrm{a} - 3}{4\left(\mathrm{a} - 1\right)}\right) \\ = \frac{3\mathrm{a} - 3 - \mathrm{a} + 3}{\left(\mathrm{a} - 1\right)} = \frac{2\mathrm{a}}{\left(\mathrm{a} - 1\right)} \end{gathered}$$

A.

0

$$\begin{gathered} \mathrm{We} \mathrm{have}, \\ \angle \mathrm{A} + \angle \mathrm{C} = 180° \\ \angle \mathrm{B} + \angle \mathrm{C} = 180° \\ \mathrm{Now}, \cos \left(\mathrm{A}\right) + \cos \left(\mathrm{C}\right) + \cos \left(\mathrm{B}\right) +\cos \left(\mathrm{D}\right) \\ =2\cos \left(\frac{\mathrm{A} + \mathrm{C}}{2}\right)\cos \left(\frac{\mathrm{A} - \mathrm{C}}{2}\right) + 2\cos \left(\frac{\mathrm{B} + \mathrm{D}}{2}\right)\cos \left(\frac{\mathrm{B} - \mathrm{D}}{2}\right) \\ =2\cos \left(90°\right)\cos \left(\frac{\mathrm{A} - \mathrm{C}}{2} \right)+ 2\cos \left(90°\right)\cos \left(\frac{\mathrm{B} - \mathrm{D}}{2}\right) = 0 \end{gathered}$$

A.

0

$$\begin{gathered} \mathrm{We} \mathrm{have}, \\ \mathrm{y} = \mathrm{Acos}\left(\mathrm{nx}\right) + \mathrm{Bsin}\left(\mathrm{nx}\right) \\ \mathrm{On} \mathrm{differentiating} \mathrm{w}.\mathrm{r}.\mathrm{t} \mathrm{x}, \mathrm{we} \mathrm{get} \\ {\mathrm{y}}_1 = - \mathrm{Ansin}\left(\mathrm{nx}\right) + \mathrm{Bncos}\left(\mathrm{nx}\right) \\ \mathrm{Again} \mathrm{differentiating} \mathrm{w}.\mathrm{r}.\mathrm{t} \mathrm{x}, \mathrm{we} \mathrm{get} \\ {\mathrm{y}}_2 = - {\mathrm{An}}^2\cos \left(\mathrm{nx}\right) - {\mathrm{Bn}}^2\sin \left(\mathrm{nx}\right) \\ \Rightarrow {\mathrm{y}}_2 = - {\mathrm{n}}^2\left(\mathrm{Acos}\left(\mathrm{nx}\right) + \mathrm{Bsin}\left(\mathrm{nx}\right)\right) \\ = - {\mathrm{n}}^2\mathrm{y} \\ \Rightarrow {\mathrm{y}}_2 + {\mathrm{n}}^2\mathrm{y} = 0 \end{gathered}$$