Maximum

$\begin{aligned} X & = \frac{1}{\color{#3D99F6}{\sin^2 \theta} + 3\sin \theta \cos \theta + \color{#3D99F6}{5 \cos^2 \theta}} \quad \quad \small \color{#3D99F6}{\sin^2 \theta + \cos^2 \theta = 1} \\ & = \frac{1}{\color{#3D99F6}{1} + 3\sin \theta \cos \theta + \color{#3D99F6}{4 \cos^2 \theta}} \\ & = \frac{1}{1 + \frac{3}{2}\left(2\sin \theta \cos \theta \right) + 2\left(2 \cos^2 \theta - 1\right) + 2} \\ & = \frac{1}{3 + \frac{3}{2}\sin (2\theta) + 2 \cos (2 \theta)} \\ & = \frac{1}{3 + \frac{5}{2}\left(\frac{3}{5} \sin (2\theta) + \frac{4}{5} \cos (2 \theta)\right)} \\ & = \frac{1}{3 + \frac{5}{2}\left(\sin (2\theta) \cos \phi + \sin \phi \cos (2 \theta)\right)} \quad \quad \small \color{#3D99F6}{\text{where }\sin \phi = \frac{4}{5}, \ \cos \phi = \frac{3}{5} \implies \phi = \tan^{-1} \frac{4}{3}} \\ & = \frac{1}{3 + \frac{5}{2} \sin \left(2 \theta + \tan^{-1} \frac{4}{3} \right)} \end{aligned}$

$X$ is maximum when the denominator $3 + \frac{5}{2} \sin \left(2 \theta + \tan^{-1} \frac{4}{3} \right)$ is minimum, and $3 + \frac{5}{2} \sin \left(2 \theta + \tan^{-1} \frac{4}{3} \right)$ is minimum when $\sin \left(2 \theta + \tan^{-1} \frac{4}{3} \right) = -1$ .

$\begin{aligned} \implies X_{max} & = \frac{1}{3 + \frac{5}{2} \left( -1 \right)} = \frac{1}{\frac{1}{2}} = \boxed{2} \end{aligned}$

In order to maximize the given expression, it suffices to minimize the denominator. Using a few trigonometric identities , $\sin^2\theta+3\sin\theta\cos\theta+5\cos^2\theta=(\sin^2\theta+\cos^2\theta)+\frac{3}{2}\cdot(2\sin\theta\cos\theta)+2\cdot(2\cos^2\theta)=1+\frac{3}{2}\sin2\theta+2(\cos2\theta+1)$ On further simplification, $\sin^2\theta+3\sin\theta\cos\theta+5\cos^2\theta=\frac{3\sin2\theta+4\cos2\theta}{2}+3$ The point of the above manipulation was to use the following lemma : $-\sqrt{a^2+b^2} \le a\sin x+b\cos x \le \sqrt{a^2+b^2}$ $\implies \frac{3\sin2\theta+4\cos2\theta}{2}+3\ge \frac{-\sqrt{3^2+4^2}}{2}+3=\frac{1}{2}$ Therefore, $\text{max}\left (\frac{1}{\sin^2\theta+3\sin\theta\cos\theta+5\cos^2\theta}\right )=\frac{1}{\text{min}\left(\sin^2\theta+3\sin\theta\cos\theta+5\cos^2\theta\right )}=\frac{1}{\frac{1}{2}}=\boxed{2}$