JEE Main 2016 — Trigonometric Ratios & Identities Question with Solution

Given

$\left|\sqrt{2{\sin }^{4}x+18{\cos }^{2}x}- \sqrt{2{\cos }^{4}x+18{\sin }^{2}x}\right|=1$

$\Rightarrow \sqrt{2{\sin }^{4}x+18{\cos }^{2}x}- \sqrt{2{\cos }^{4}x+18{\sin }^{2}x}= \pm 1$

$\Rightarrow \sqrt{2{\sin }^{4}x+18{\cos }^{2}x}= \pm 1+ \sqrt{2{\cos }^{4}x+18{\sin }^{2}x}$

By squaring both the sides, we get

$2{\sin }^{4}x+18{\cos }^{2}x=1+2{\cos }^{4}x+18{\sin }^{2}x\pm 2\sqrt{2{\cos }^{4}x+18{\sin }^{2}x}$

$\Rightarrow 2\left({\sin }^{4}x-{\cos }^{4}x\right)+18\left({\cos }^{2}x-{\sin }^{2}x\right)=1\pm 2\sqrt{2{\cos }^{4}x+18{\sin }^{2}x}$

$\Rightarrow 2\left({\sin }^{2}x-{\cos }^{2}x\right)+18\left({\cos }^{2}x-{\sin }^{2}x\right)=1\pm 2\sqrt{2{\cos }^{4}x+18{\sin }^{2}x}$

$\Rightarrow 16\left({\cos }^{2}x-{\sin }^{2}x\right)=1\pm 2\sqrt{2{\cos }^{4}x+18{\sin }^{2}x}$

$\Rightarrow 16\cos 2x-1=\pm 2\sqrt{2{\left(\frac{1+\cos 2x}{2}\right)}^2+9\left(1-\cos 2x\right)}$

Squaring both sides again, we get

$256{\cos }^{2}2x+1-32\cos 2x=4\left(\frac{1+2\cos 2x+{\cos }^{2}2x}{2}+9\left(1-\cos 2x\right)\right)$

$256{\cos }^{2}2x+1-32\text{cos}2x=2\left(19-16\cos 2x+{\cos }^{2}2x\right)$

$\Rightarrow 254{\cos }^{2}2x=37\Rightarrow {\cos }^{2}2x=\frac{37}{254}\Rightarrow \cos 2x=\pm \sqrt{\frac{37}{254}}\in \left[-1, 1\right]$
Since, the solutions lie in all four quadrants, there are $8$ solutions which satisfy the equation.