Trigonometric exercise level 2

Exercise:

Solve in $\mathbb{R},$ $\large{{{\left( \sqrt[4]{17-12\sqrt{2}} \right)}^{\sqrt{\sin 2x}+\frac{1}{\sqrt{\sin 2x}}}}+{{\left( \sqrt[4]{17+12\sqrt{2}} \right)}^{\sqrt{\sin 2x}+\frac{1}{\sqrt{\sin 2x}}}}=6}$

Solution: we have $\sqrt[4]{17+12\sqrt{2}}=\sqrt[4]{17+2\left( 3 \right)\left( 2\sqrt{2} \right)}=\sqrt[4]{9+8+2\left( 3 \right)\left( 2\sqrt{2} \right)}=\sqrt[4]{{{\left( 3+2\sqrt{2} \right)}^{2}}}=\sqrt{3+2\sqrt{2}}$

And $\sqrt[4]{17-12\sqrt{2}}=\sqrt[4]{{{\left( 3-2\sqrt{2} \right)}^{2}}}=\sqrt{3-2\sqrt{2}}$

Notice that $\frac{1}{\sqrt{3-2\sqrt{2}}}\times \frac{\sqrt{3+2\sqrt{2}}}{\sqrt{3+2\sqrt{2}}}=\frac{\sqrt{3+2\sqrt{2}}}{9-8}=\sqrt{3+2\sqrt{2}}$

Hence $\sqrt{3+2\sqrt{2}}=\frac{1}{\sqrt{3-2\sqrt{2}}}={{\left( \sqrt{3-2\sqrt{2}} \right)}^{-1}}$

Thus $\large{{{\left( \sqrt{3-2\sqrt{2}} \right)}^{\sqrt{\sin 2x}+\frac{1}{\sqrt{\sin 2x}}}}+{{\left( \sqrt{3-2\sqrt{2}} \right)}^{-\left( \sqrt{\sin 2x}+\frac{1}{\sqrt{\sin 2x}} \right)}}=6}$

Take $\sqrt{3-2\sqrt{2}}=t\Leftrightarrow {{t}^{y}}+{{t}^{-y}}=6$ where $y=\sqrt{\sin 2x}+\frac{1}{\sqrt{\sin 2x}}$

So ${{t}^{y}}+{{t}^{-y}}={{t}^{y}}+\frac{1}{{{t}^{y}}}=6\Leftrightarrow {{t}^{2y}}+1=6{{t}^{y}}\Leftrightarrow {{\left( {{t}^{y}} \right)}^{2}}-6{{t}^{y}}+1=0$

${{\left( {{t}^{y}} \right)}^{2}}-2\left( 3 \right){{t}^{y}}+9-9+1=0\Leftrightarrow {{\left( {{t}^{y}}-3 \right)}^{2}}-8=0\Leftrightarrow {{\left( {{t}^{y}}-3 \right)}^{2}}=8\Leftrightarrow {{t}^{y}}=3\pm 2\sqrt{2}$

So if $y=1\Leftrightarrow \sqrt{\sin 2x}+\frac{1}{\sqrt{\sin 2x}}=1$ notice that $f\left( x \right)=\sqrt{\sin 2x}+\frac{1}{\sqrt{\sin 2x}}$ is defined when $\sin 2x>0$ so $u\left( x \right)=\sin 2x$ belong to the first and second quadrant hence $2x\in \left[ 0,\pi  \right]$

Thus $x\in \left[ 0,\frac{\pi }{2} \right]$

If ${{t}^{y}}=3-2\sqrt{2}\Leftrightarrow {{t}^{y}}={{t}^{2}}\Leftrightarrow y=2$

If ${{t}^{y}}=3+2\sqrt{2}=\frac{1}{3-2\sqrt{2}}={{t}^{-2}}\Leftrightarrow y=-2$  rejected since $y>0$

So $\sqrt{u\left( x \right)}+\frac{1}{\sqrt{u\left( x \right)}}=2\Leftrightarrow u\left( x \right)+1=2\sqrt{u\left( x \right)}\Leftrightarrow {{u}^{2}}\left( x \right)+1+2u\left( x \right)=4u\left( x \right)$

$\Leftrightarrow {{u}^{2}}\left( x \right)-2u\left( x \right)+1=0\Leftrightarrow {{\left( u\left( x \right)-1 \right)}^{2}}=0\Leftrightarrow u\left( x \right)=1\Leftrightarrow \sin 2x=1$

$\Leftrightarrow \sin 2x=\sin \left( \frac{\pi }{2} \right)\Leftrightarrow 2x=\frac{\pi }{2}+2k\pi \,\,or\,\,\,\pi -2x=\frac{\pi }{2}+2k\pi \,\,\,\,,\,\,\,k\in \mathbb{Z}$

So $x=\frac{\pi }{4}+k\pi \,\,\,\,or\,\,\,\,\,x=\frac{\pi }{4}-k\pi \,\,\,\,\,,\,\,k\in \mathbb{Z}$ since $x\in \left[ 0,\frac{\pi }{2} \right]$ So $x=\frac{\pi }{4}$ is accepted value