The number of x ∈ [ 0, 2 π] for which | 2 ^ 4 x + 18 ^ 2 x - 2 ^ 4...

Given

\left|\sqrt{2\sin^{4} x + 18\cos^{2} x} - \sqrt{2\cos^{4} x + 18\sin^{2} x}\right| = 1$<br/><br/>$⇒ \sqrt{2\sin^{4} x + 18\cos^{2} x} - \sqrt{2\cos^{4} x + 18\sin^{2} x} = \pm 1$<br/><br/>$⇒ \sqrt{2\sin^{4} x + 18\cos^{2} x} = \pm 1 + \sqrt{2\cos^{4} x + 18\sin^{2} x}$<br/><br/>By squaring both the sides, we get<br/><br/>$2\sin^{4} x + 18\cos^{2} x = 1 + 2\cos^{4} x + 18\sin^{2} x \pm 2\sqrt{2\cos^{4} x + 18\sin^{2} x}$<br/><br/>$⇒ 2\left(\sin^{4} x - \cos^{4} x\right) + 18\left(\cos^{2} x - \sin^{2} x\right) = 1 \pm 2\sqrt{2\cos^{4} x + 18\sin^{2} x}$<br/><br/>$⇒ 2\left(\sin^{2} x - \cos^{2} x\right) + 18\left(\cos^{2} x - \sin^{2} x\right) = 1 \pm 2\sqrt{2\cos^{4} x + 18\sin^{2} x}$<br/><br/>$⇒ 16\left(\cos^{2} x - \sin^{2} x\right) = 1 \pm 2\sqrt{2\cos^{4} x + 18\sin^{2} x}$<br/><br/>$⇒ 16\cos 2x - 1 = \pm 2\sqrt{2{\left(\dfrac{1 + \cos 2x}{2}\right)}^{2} + 9\left(1 - \cos 2x\right)}$<br/><br/>Squaring both sides again, we get<br/><br/>$256\cos^{2} 2x + 1 - 32\cos 2x = 4 \left(\dfrac{1 + 2\cos 2x + \cos^{2} 2x}{2} ++9 \left(1 - \cos 2x\right)\right)

$256\cos^{2} 2x + 1 -$32\text\cos2x = 2 \left(19 - 16\cos 2x + \cos^{2} 2x\right)$<br/><br/>$⇒ 254\cos^{2} 2x = 37 ⇒ \cos^{2} 2x = \dfrac{37}{254}⇒ \cos 2x = $\pm$\sqrt{\dfrac{37}{254}} \in \left[-1, 1\right]
Since, the solutions lie in all four quadrants, there are $8$ solutions which satisfy the equation.