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The number of $x \in[0,2 \pi]$ for which $\left|\sqrt{2 \sin ^{4} x+18 \cos ^{2} x}-\sqrt{2 \cos ^{4} x+18 \sin ^{2} x}\right|=1$ is:
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\begin{aligned} &\sin ^{2} x+\cos ^{2} x=1 \\ &\sin ^{4} x+\cos ^{4} x+2 \sin ^{2} x \cos ^{2} x=1 \\ &\sin ^{4} x+\cos ^{4} x=1-2 \sin ^{2} x \cos ^{2} x \quad \ldots[1] \\ &\sin ^{2} x+\cos ^{2} x=1 \\ &\sin ^{6} x+\cos ^{6} x+3 \sin ^{2} x \cos ^{2} x\left(\sin ^{2} x+\cos ^{2} x\right)=1 \\ &\sin ^{6} x+\cos ^{6} x=1-3 \sin ^{2} x \cos ^{2} x \quad \ldots[2] \\ &\mid \sqrt{2 \sin ^{4} x+18 \cos ^{2} x}-\sqrt{2 \cos ^{4} x+18 \sin ^{2} x}=1 \end{aligned}
Squaring above equation
\begin{aligned} &\therefore 2 \sin ^{4} x+18 \cos ^{2} x+2 \cos ^{4} x+18 \sin ^{2} x- \\ &2\left(\sqrt{2 \sin ^{4} x+18 \cos ^{2} x}\right)\left(\sqrt{2 \cos ^{4} x+18 \sin ^{2} x}\right)=1 \\ &\therefore 2\left(\sin ^{4} x+\cos ^{4} x\right)+18\left(\cos ^{2} x+\sin ^{2} x\right)-1= \\ &2\left(\sqrt{2 \sin ^{4} x+18 \cos ^{2} x}\right)\left(\sqrt{2 \cos ^{4} x+18 \sin ^{2} x}\right) \end{aligned}

$\therefore 2\left(1-2 \sin ^{2} x \cos ^{2} x\right)+18-1=2\left(\sqrt{2 \sin ^{4} x+18 \cos ^{2} x}\right)\left(\sqrt{2 \cos ^{4} x+18 \sin ^{2} x}\right)$
$. .($ [Using[1] )
$\therefore 19-4 \sin ^{2} x \cos ^{2} x=2\left(\sqrt{2 \sin ^{4} x+18 \cos ^{2} x}\right)\left(\sqrt{2 \cos ^{4} x+18 \sin ^{2} x}\right)$
Squaring above equation
$\therefore 361+16 \sin ^{4} x \cos ^{4} x-152 \sin ^{2} x \cos ^{2} x=4\left(4 \sin ^{4} x \cos ^{4} x+36 \cos ^{6} x+36 \sin ^{6} x+\right.$
$324 \sin ^{2} x \cos ^{2} x$ )
$\therefore 361+16 \sin ^{4} x \cos ^{4} x-152 \sin ^{2} x \cos ^{2} x=16 \sin ^{4} x \cos ^{4} x+144 \cos ^{6} x+144 \sin ^{6} x+$ $1296 \sin ^{2} x \cos ^{2} x$
$\therefore 361=144\left(\cos ^{6} x+\sin ^{6} x\right)+1448 \sin ^{2} x \cos ^{2} x$
$\therefore 361=144\left(1-3 \sin ^{2} x \cos ^{2} x\right)+1448 \sin ^{2} x \cos ^{2} x$
(Using [2] )
$\therefore 217=1016 \sin ^{2} x \cos ^{2} x$

\begin{aligned}
&\therefore 217=1016 \sin ^{2} \mathrm{x} \cos ^{2} \mathrm{x} \\
&\therefore 217=254 \sin ^{2} 2 \mathrm{x} \\
&\therefore \sin 2 \mathrm{x}=\pm \sqrt{\frac{217}{254}} \\
&\text { Let } \sin ^{-1} \sqrt{\frac{217}{254}}=\theta \\
&\sin 2 \mathrm{x}=\pm \sqrt{\frac{217}{254}} \\
&\therefore 2 \mathrm{x}=\sin ^{-1} \sqrt{\frac{217}{254}} \\
&\therefore 2 \mathrm{x}=\theta, \pi-\theta, 2 \pi+\theta, 3 \pi-\text { theta } \\
&\therefore \mathrm{x}=\frac{\pi}{2}, \frac{\pi-\theta}{2}, \frac{2 \pi+\theta}{2}, \frac{3 \pi-\theta}{2}
\end{aligned}

$2 x=\sin ^{-1}\left(-\sqrt{\frac{217}{254}}\right)$
$\therefore 2 \mathrm{x}=\pi+\theta, 2 \pi-\theta, 3 \pi+\theta, 4 \pi-\theta$
$\therefore \mathrm{x}=\frac{\pi+\theta}{2}, \frac{2 \pi-\theta}{2}, \frac{3 \pi+\theta}{2}, \frac{4 \pi-\theta}{2}$
So, total number of solutions $=8$
by Platinum (122,442 points)

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