Question

Find all numbers $x,y$that satisfy the equation

${({\sin }^{2}x+\frac{1}{{\sin }^{2}x})}^2+{({\cos }^{2}x+\frac{1}{{\cos }^{2}x})}^2=12+\frac{1}{2}\sin y$

Answer 1

Using $R.M.S\ge A.M$
$\sqrt{\frac{{(si{n}^{2}x+\frac{1}{si{n}^{2}x})}^2+{(co{s}^{2}x+\frac{1}{co{s}^{2}x})}^2}{2}}\ge \frac{si{n}^{2}x+\frac{1}{si{n}^{2}x}+co{s}^{2}x+\frac{1}{co{s}^{2}x}}{2}$
$\Rightarrow \sqrt{\frac{{(si{n}^{2}x+\frac{1}{si{n}^{2}x})}^2+{(co{s}^{2}x+\frac{1}{co{s}^{2}x})}^2}{2}}\ge \frac{1+\frac{4}{{\sin }^{2}2x}}{2}$
$\Rightarrow \sqrt{\frac{{(si{n}^{2}x+\frac{1}{si{n}^{2}x})}^2+{(co{s}^{2}x+\frac{1}{co{s}^{2}x})}^2}{2}}\ge \frac{5}{2}$
$\Rightarrow {(si{n}^{2}x+\frac{1}{si{n}^{2}x})}^2+{(co{s}^{2}x+\frac{1}{co{s}^{2}x})}^2\ge \frac{25}{2}$
And as $12+\frac{1}{2}\sin y\le \frac{25}{2}$
Hence ${(si{n}^{2}x+\frac{1}{si{n}^{2}x})}^2+{(co{s}^{2}x+\frac{1}{co{s}^{2}x})}^2=\frac{25}{2}$ and $12+\frac{1}{2}\sin y=\frac{25}{2}$
$\Rightarrow {\sin }^{2}x=\frac{1}{2}$ or $\sin y=1$
$\Rightarrow x=n\pi \pm \frac{\pi }{4}$ or $y=(4m+1)\frac{\pi }{2}$