Find the real root for this equation

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

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 the accepted value

Assuming that $(a \pm b \sqrt 2)^4 = 17 \pm 12\sqrt 2$ $\implies \sqrt [4] {17 \pm 12\sqrt 2} = a \pm b \sqrt 2$ .

$\begin{aligned} (a \pm b \sqrt 2)^4 & = 17 \pm 12\sqrt 2 \\ a^4 \pm 4a^3b\sqrt 2 + 12a^2b^2 \pm 8ab^3\sqrt 2 + 4b^4 & = 17 \pm 12\sqrt 2 \end{aligned}$

Equating the rational and irrational parts, we have:

$\begin{cases} a^4 + 12a^2b^2 + 4b^4 = 17 \\ 4a^3b + 8ab^3 = 12 \implies a^3b + 2ab^3 = 3 \end{cases}$

By observation, we note that $a = b = 1$ . $\implies \sqrt [4] {17 \pm 12\sqrt 2} = 1 \pm \sqrt 2$ .

Let $y = \sqrt{\sin 2x} + \dfrac 1{\sqrt{\sin 2x}}$ . Then we have:

\begin{aligned} X & = \left(\sqrt [4] {17 - 12\sqrt 2}\right)^y + \left(\sqrt [4] {17 + 12\sqrt 2}\right)^y \\ & = \left(1-\sqrt 2 \right)^\color{#3D99F6}{y} + \left(1+\sqrt 2 \right)^\color{#3D99F6}{y} & \small \color{#3D99F6}{\text{By observation }y=2} \\ & = \left(1-\sqrt 2 \right)^\color{#3D99F6}{2} + \left(1+\sqrt 2 \right)^\color{#3D99F6}{2} \\ & = 1 - 2\sqrt 2 + 2 + 1 + 2\sqrt 2 + 2 \\ & = 6 \end{aligned}

$\begin{aligned} \implies \color{#3D99F6}{\sqrt{\sin 2x}} + \dfrac 1{\color{#3D99F6}{\sqrt{\sin 2x}}} & = 2 & \small \color{#3D99F6}{\text{Let }u = \sqrt{\sin 2x}} \\ u + \frac 1u & = 2 \\ u^2 - 2u + 1 & = 0 \\ (u-1)^2 & = 0 \\ u & = 1 \\ \implies \sqrt{\sin 2x} & = 1 \\ \sin 2x & = 1 \\ \implies x & = \frac \pi 4 \approx \boxed{0.785} \end{aligned}$