A.M G.M?

Update: New solution added :)

Solution 1: Let $f(\theta) = \sin \theta + \frac{2}{\sin \theta}$ Then for minimization, we must have $f'(\theta) = \cos \theta - \frac{2}{\sin ^2 \theta} \cdot \cos \theta = \cos \theta \left( 1 - \frac{2}{\sin ^2 \theta} \right) = 0$ Then, under the domain $(0, \pi)$ ,

Case 1 $\cos \theta = 0 \implies \theta = \frac{\pi}{2}$ Case 2 (not possible) $1 - \frac{2}{\sin ^2 \theta} = 0 \implies \sin ^2 \theta = 2 \ge 1$

As $\theta \rightarrow 0^+$ and $\theta \rightarrow \pi^-$ , the value of $f$ approaches $+\infty$ . Hence, the minimum value occurs at $\theta = \frac{\pi}{2}$ , and $f\left( \frac{\pi}{2}\right) = \sin \frac{\pi}{2} + \frac{2}{\sin \frac{\pi}{2}}=\fbox{3}$

Solution 2: @Raj Rajput @Calvin Lin I think Titu's lemma also works for this problem, since $\theta \in (0,\pi)$ , and thus $\sin \theta > 0$ . We have $\sin \theta + \frac{2}{\sin \theta} = \frac{\sin \theta}{1} + \frac{1}{\sin \theta} + \frac{1}{\sin \theta} \geqslant \frac{(\sin \theta + 1)^2}{1 + \sin \theta} + \frac{1}{\sin \theta} = \sin \theta + 1 + \frac{1}{\sin \theta}$ Equality occurs when $\sin \theta + \frac{2}{\sin \theta} = \sin \theta + 1 + \frac{1}{\sin \theta} \Longrightarrow \frac{1}{\sin \theta} =1 \Longrightarrow \sin \theta = 1 \Longrightarrow \theta = \frac{\pi}{2}$ Therefore, the minimum value is $\sin \frac{\pi}{2} + \frac{2}{\sin \frac{\pi}{2}} = \boxed{3}$ .

I think that the title of this question was little bit misleading becase this question could be done by simple algebraic manipulation.

Y= Sin X+2/(Sin X)=Sin X (Sin X × Sin X +2)

As we know that Sin value can be maximum 1 , therefore y(max)=3.

$0 < \theta < π$

$0 < \sin\theta ≤ 1$

$\sin\theta + \frac{2}{\sin\theta}$

$= ( \sin\theta + \frac{1}{\sin\theta}) + \frac{1}{\sin\theta}$

When $\sin\theta = 1$ ,

$\sin\theta + \frac{1}{\sin\theta}$ has minimum value $2$ and $\frac{1}{\sin\theta}$ has minimum value $1$ at the same time.

So the minimum value of $\sin\theta + \frac{2}{\sin\theta}$ is $2 + 1 = \boxed{3}$

It should be when sin x is a positive real no., for all the real values of x, this equation has min. value of -(infinity). thus , the correct option should be none of these..... would be much easier then :)