Equation exercise solved by using AM-GM inequality asked in arabic math group

Exercise:

Solve in $\mathbb{R}$ , $\frac{{{11}^{\cos x}}}{{{11}^{\sin x}}}+\frac{{{11}^{\sin x}}}{{{11}^{\cos x}}}=\sqrt{2}\left( \cos x+\sin x \right)$

Solution: Let $w=\frac{{{11}^{\cos x}}}{{{11}^{\sin x}}}\Leftrightarrow \frac{1}{w}=\frac{{{11}^{\sin x}}}{{{11}^{\cos x}}}$

So $w+\frac{1}{w}=\sqrt{2}\left( \cos x+\sin x \right)=\sqrt{2}\left( \frac{\sqrt{2}}{\sqrt{2}}\cos x+\frac{\sqrt{2}}{\sqrt{2}}\sin x \right)$

$\Leftrightarrow w+\frac{1}{w}=2\left( \sin \frac{\pi }{4}\cos x+\cos \frac{\pi }{4}\sin x \right)=2\sin \left( \frac{\pi }{4}+x \right)$

But $AM \ge GM$ hence $w+\frac{1}{w}\ge 2\sqrt{w\left( \frac{1}{w} \right)}=2$

So $2\le w+\frac{1}{w}\Leftrightarrow 2\le 2\sin \left( \frac{\pi }{4}+x \right)\Leftrightarrow \sin \left( \frac{\pi }{4}+x \right)\ge 1$ ( 1 )

But $\left| \cos x+\sin x \right|=\sqrt{2}\left| \sin \left( \frac{\pi }{4}+x \right) \right|\le \sqrt{2}$

$\Leftrightarrow \left| \sqrt{2}\left( \cos x+\sin x \right) \right|=\sqrt{2}\left| \cos x+\sin x \right|\le 2\Leftrightarrow \sin \left( \frac{\pi }{4}+x \right)\le 1$  ( 2 )

By (1) and (2) we get $\sin \left( \frac{\pi }{4}+x \right)=1$ hence $x=\frac{\pi }{4}+2k\pi \,\,\,or\,\,x=\frac{5\pi }{4}+2k\pi \,\,\,\,,\,\,k\in \mathbb{Z}$

So $w+\frac{1}{w}=2\Leftrightarrow {{w}^{2}}-2w+1=0\Leftrightarrow w=1$ (double roots)

Hence ${{11}^{\cos x}}={{11}^{\sin x}}\Leftrightarrow \cos x=\sin x\Leftrightarrow \tan x=1=\tan \left( \frac{\pi }{4} \right)\Leftrightarrow x=\frac{\pi }{4}+k\pi $



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Idea of AM-GM credit to Amer Khamiseh    &   Utsabraj Sarkar