To find minimum value of $2\csc(2x)+\sec(x)+\csc(x)$ To find minimum value of $$2\csc(2x)+\sec(x)+\csc(x)$$ for $x \in (0,\frac{\pi}{2})$
i converted to sin and cos so that i may reduce it to form $a\cos(x)+b\sin(x)$. I reduced it to $$\frac{\sin(x)+\cos(x)+1}{\sin(x)\cos(x)}$$
How do i proceed from here?
Thanks a ton!
 A: Method $\#1:$ 
If we set $\sin x+\cos x=u, u\le\sqrt2, u^2=1+2\sin x\cos x$
we need to minimize $$2\cdot\dfrac{u+1}{u^2-1}=\dfrac2{u-1}$$
i.e.,  to maximize $u-1$
Method $\#2:$
As $(\sin x+\cos x)^2-1^2=2\sin x\cos x$
and $\sin x+\cos x=\sqrt2\sin\left(x+\dfrac\pi4\right)$
$$\dfrac{\sin x+\cos x+1}{\sin x\cos x}=\dfrac2{\sin x+\cos x-1}=\dfrac2{\sqrt2\sin\left(x+\dfrac\pi4\right)-1}$$
Now for $x\in\left(0,\dfrac\pi2\right),$
$$\dfrac1{\sqrt2}<\sin\left(x+\dfrac\pi4\right)\le1$$
Method $\#3:$
$$2\sin x\cos x=-\cos2\left(\dfrac\pi4+x\right)=\left(\sqrt2\sin\left(x+\dfrac\pi4\right)\right)^2-1^2$$
A: Using $\sin2A=2\sin A\cos A,\cos2A=\cos^2A-\sin^2A$
$$\dfrac{\sin x+\cos x+1}{\sin x\cos x}=\dfrac{2\cos\dfrac x2\left(\cos\dfrac x2+\sin\dfrac x2\right)}{2\cos\dfrac x2\sin\dfrac x2\left(\cos^2\dfrac x2-\sin^2\dfrac x2\right)}=\dfrac1{\sin\dfrac x2\left(\cos\dfrac x2-\sin\dfrac x2\right)}$$
Now $\cos\dfrac x2-\sin\dfrac x2=\sqrt2\sin\left(\dfrac\pi4-\dfrac x2\right)$
$\implies\sin\dfrac x2\left(\cos\dfrac x2-\sin\dfrac x2\right)=\sqrt2\sin\dfrac x2\sin\left(\dfrac\pi4-\dfrac x2\right)=\dfrac{\cos\left(x-\dfrac\pi8\right)-\cos\dfrac\pi8}{\sqrt2}$ which needs to be maximized with $x \in \left(0,\dfrac\pi2\right)$
