# To find minimum value of 2 csc &#x2061;<!-- ⁡ --> ( 2 x ) + sec &#x20

To find minimum value of $2\mathrm{csc}\left(2x\right)+\mathrm{sec}\left(x\right)+\mathrm{csc}\left(x\right)$
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Method #1:
If we set $\mathrm{sin}x+\mathrm{cos}x=u,u\le \sqrt{2},{u}^{2}=1+2\mathrm{sin}x\mathrm{cos}x$
we need to minimize
$2\cdot \frac{u+1}{{u}^{2}-1}=\frac{2}{u-1}$
i.e., to maximize u−1
Method #2:
As $\left(\mathrm{sin}x+\mathrm{cos}x{\right)}^{2}-{1}^{2}=2\mathrm{sin}x\mathrm{cos}x$
and $\mathrm{sin}x+\mathrm{cos}x=\sqrt{2}\mathrm{sin}\left(x+\frac{\pi }{4}\right)$
$\frac{\mathrm{sin}x+\mathrm{cos}x+1}{\mathrm{sin}x\mathrm{cos}x}=\frac{2}{\mathrm{sin}x+\mathrm{cos}x-1}=\frac{2}{\sqrt{2}\mathrm{sin}\left(x+\frac{\pi }{4}\right)-1}$
Now for $x\in \left(0,\frac{\pi }{2}\right),$,
$\frac{1}{\sqrt{2}}<\mathrm{sin}\left(x+\frac{\pi }{4}\right)\le 1$
Method #3:
$2\mathrm{sin}x\mathrm{cos}x=-\mathrm{cos}2\left(\frac{\pi }{4}+x\right)={\left(\sqrt{2}\mathrm{sin}\left(x+\frac{\pi }{4}\right)\right)}^{2}-{1}^{2}$