Explanation

(c)
Given, $\left( 5 + 4\cos\theta \right)\left( 2\cos{\theta + 1} \right) = 0$ …(i)
Since, $\cos{\theta = \frac{1 - \tan^{2}\frac{\theta}{2}}{1 + \tan^{2}\frac{\theta}{2}}} = \frac{1 - t^{2}}{1 + t^{2}}$ $\left\lbrack put\tan{\frac{\theta}{2} = t} \right\rbrack$
$\therefore$ From Eq. (i),
$\left\lbrack 5 + 4\left( \frac{1 - t^{2}}{1 - t^{2}} \right) \right\rbrack\left\lbrack 2\left( \frac{1 - t^{2}}{1 + t^{2}} \right) + 1 \right\rbrack = 0$
$\Rightarrow \left\lbrack 5 + 5t^{2} + 4 - 4t^{2} \right\rbrack\left\lbrack 2 - 2t^{2} + 1 + t^{2} \right\rbrack = 0$
$\Rightarrow \left( t^{2} + 9 \right)\left( 3 - t^{2} \right) = 0\ \ \Rightarrow \ \ t = \pm \sqrt{3}$
$\therefore\ \tan{\frac{\theta}{2} = \sqrt{3}}$ or $\tan{\frac{\theta}{2} = - \sqrt{3}}$
$\Rightarrow \frac{\theta}{2} = \frac{\pi}{3}$ or $\frac{\theta}{2} = \frac{2\pi}{3}$
$\therefore\ \ \ \theta = \frac{2\pi}{3}$ or $\frac{4\pi}{3}$