Verified answer

Solution:

[tex]\begin{aligned}&\left|\sqrt{2\sin^2x + 18 \cos^2x} - \sqrt{2\cos^2x + 18 \sin^2x} \right| = 1\\\implies& \left|\sqrt{2\sin^2x + 2 \cos^2x + 16\cos^2x} - \sqrt{2\cos^2x + 2 \sin^2x + 16\sin^2x} \right| = 1\\\implies& \left|\sqrt{2(\sin^2x + \cos^2x) + 16\cos^2x} - \sqrt{2(\cos^2x + \sin^2x) + 16\sin^2x} \right| = 1\\\implies& \left|\sqrt{2+ 16\cos^2x} - \sqrt{2 + 16\sin^2x} \right| = 1\end{aligned}[/tex]

Squaring both sides we get,

[tex]\begin{aligned}& (2+ 16\cos^2x) + (2 + 16\sin^2x) - 2\sqrt{(2+ 16\cos^2x) (2 + 16\sin^2x)} = 1\\\implies& 4 + 16(\sin^2x + \cos^2x) - 2\sqrt{2\cdot 2 \cdot (1+ 8\cos^2x) (1+ 8\sin^2x)} = 1\\\implies &20 - 4\sqrt{1+8\cos^2x + 8\sin^2x + 8^2 \sin^2x \cos^2x } = 1\\\implies &19 = 4\sqrt{1+8(\cos^2x + \sin^2x) + 16 \cdot (2\sin x \cos x)^2 }\\\implies &\frac{19}{4} = \sqrt{9 + 16 \cdot (2\sin x \cos x)^2 }\\\implies &\frac{19}{4} = \sqrt{9 + 16 \sin^2(2x) }\end{aligned}[/tex]

Again squaring both sides,

[tex]\begin{aligned}& \frac{361}{16} = 9 + 16 \sin^2(2x)\\\implies& \frac{361}{16} - 9 = 16 \sin^2(2x)\\\implies &\frac{361 - 144}{16} = 16 \sin^2(2x)\\\implies& \frac{217}{256} = \sin^2(2x)\\\implies& \pm\frac{\sqrt{217}}{16} = \sin(2x)\end{aligned}[/tex]

Taking the positive version first,

[tex]\begin{aligned}&\frac{\sqrt{217}}{16} = \sin(2x)\\\implies & \sin^{-1}\left(\dfrac{\sqrt{217}}{16}\right) = 2x\\\implies &\dfrac12\sin^{-1}\left(\dfrac{\sqrt{217}}{16}\right) = x \end{aligned}[/tex]

Now, all the solutions in the interval [tex][0, 2\pi][/tex] are given by,

[tex]\begin{aligned}x&= \dfrac12 \sin^{-1}\left(\frac{\sqrt{217}}{16}\right)\\ x & = \frac\pi2 + \dfrac12 \sin^{-1}\left(\frac{\sqrt{217}}{16}\right)\\ x & = \pi + \dfrac12 \sin^{-1}\left(\frac{\sqrt{217}}{16}\right)\\ x & = \frac{3\pi}2 + \dfrac12 \sin^{-1}\left(\frac{\sqrt{217}}{16}\right)\end{aligned}[/tex]

Now taking the negative version,

[tex]\begin{aligned}&-\frac{\sqrt{217}}{16} = \sin(2x)\\\implies & \sin^{-1}\left(-\dfrac{\sqrt{217}}{16}\right) = 2x\\\implies &-\dfrac12\sin^{-1}\left(\dfrac{\sqrt{217}}{16}\right) = x \end{aligned}[/tex]

Now, all the solutions in the interval [tex][0, 2\pi][/tex] are given by,

[tex]\begin{aligned}x&= \frac\pi 2- \dfrac12 \sin^{-1}\left(\frac{\sqrt{217}}{16}\right)\\ x & = \pi - \dfrac12 \sin^{-1}\left(\frac{\sqrt{217}}{16}\right)\\ x & = \frac{3\pi} 2- \dfrac12 \sin^{-1}\left(\frac{\sqrt{217}}{16}\right)\\ x & = 2\pi - \dfrac12 \sin^{-1}\left(\frac{\sqrt{217}}{16}\right)\end{aligned}[/tex]

Hence these are total of 8 solutions.

Cheers!