Question

The minimum value of $\alpha$ for which the equation $\frac{4}{\sin x}+\frac{1}{1-\sin x}=\alpha$ has at least one solution in $(0,\frac{\pi }{2})$ is

Answer 1

Let $f\left(x\right)=\frac{4}{\sin x}+\frac{1}{1-\sin x}$ and $\sin x=t$
$\because x\in (0,\frac{\pi }{2})\Rightarrow 0<t<1$
$f\left(x\right)=\frac{4}{t}+\frac{1}{1-t}$
$f^{\prime }\left(x\right)=\frac{-4}{t^2}+\frac{1}{(1-t{)}^2}=0$
$\Rightarrow \frac{t^2-4(1-t{)}^2}{t^2(1-t{)}^2}=0$
$\Rightarrow t=\frac{2}{3}$
$f_{\text{min}}$ at $t=\frac{2}{3}$
${\alpha }_{\text{min}}=f\left(\frac{2}{3}\right)=\frac{4}{2/3}+\frac{1}{1-2/3}$
$=6+3$
$=9$