JEE Main 2016 — Mathematics Trigonometry

Given

$$|\sqrt{2{\mathrm{sin}}^{4}⁡x+18{\mathrm{cos}}^{2}⁡x}- \sqrt{2{\mathrm{cos}}^{4}⁡x+18{\mathrm{sin}}^{2}⁡x}|=1$$∣2sin4⁡x+18cos2⁡x​−2cos4⁡x+18sin2⁡x​∣=1

$$\Rightarrow \sqrt{2{\mathrm{sin}}^{4}⁡x+18{\mathrm{cos}}^{2}⁡x}- \sqrt{2{\mathrm{cos}}^{4}⁡x+18{\mathrm{sin}}^{2}⁡x}= \pm 1$$⇒2sin4⁡x+18cos2⁡x​−2cos4⁡x+18sin2⁡x​=±1

$$\Rightarrow \sqrt{2{\mathrm{sin}}^{4}⁡x+18{\mathrm{cos}}^{2}⁡x}= \pm 1+ \sqrt{2{\mathrm{cos}}^{4}⁡x+18{\mathrm{sin}}^{2}⁡x}$$⇒2sin4⁡x+18cos2⁡x​=±1+2cos4⁡x+18sin2⁡x​

By squaring both the sides, we get

$$2{\mathrm{sin}}^{4}⁡x+18{\mathrm{cos}}^{2}⁡x=1+2{\mathrm{cos}}^{4}⁡x+18{\mathrm{sin}}^{2}⁡x\pm 2\sqrt{2{\mathrm{cos}}^{4}⁡x+18{\mathrm{sin}}^{2}⁡x}$$2sin4⁡x+18cos2⁡x=1+2cos4⁡x+18sin2⁡x±22cos4⁡x+18sin2⁡x​

$$\Rightarrow 2({\mathrm{sin}}^{4}⁡x-{\mathrm{cos}}^{4}⁡x)+18({\mathrm{cos}}^{2}⁡x-{\mathrm{sin}}^{2}⁡x)=1\pm 2\sqrt{2{\mathrm{cos}}^{4}⁡x+18{\mathrm{sin}}^{2}⁡x}$$⇒2(sin4⁡x−cos4⁡x)+18(cos2⁡x−sin2⁡x)=1±22cos4⁡x+18sin2⁡x​

$$\Rightarrow 2({\mathrm{sin}}^{2}⁡x-{\mathrm{cos}}^{2}⁡x)+18({\mathrm{cos}}^{2}⁡x-{\mathrm{sin}}^{2}⁡x)=1\pm 2\sqrt{2{\mathrm{cos}}^{4}⁡x+18{\mathrm{sin}}^{2}⁡x}$$⇒2(sin2⁡x−cos2⁡x)+18(cos2⁡x−sin2⁡x)=1±22cos4⁡x+18sin2⁡x​

$$\Rightarrow 16({\mathrm{cos}}^{2}⁡x-{\mathrm{sin}}^{2}⁡x)=1\pm 2\sqrt{2{\mathrm{cos}}^{4}⁡x+18{\mathrm{sin}}^{2}⁡x}$$⇒16(cos2⁡x−sin2⁡x)=1±22cos4⁡x+18sin2⁡x​

$$\Rightarrow 16\mathrm{cos}⁡2x-1=\pm 2\sqrt{2{(\frac{1+\mathrm{cos}⁡2x}{2})}^{2}+9(1-\mathrm{cos}⁡2x)}$$⇒16cos⁡2x−1=±22(21+cos⁡2x​)2+9(1−cos⁡2x)​

Squaring both sides again, we get

$$256{\mathrm{cos}}^{2}⁡2x+1-32\mathrm{cos}⁡2x=4(\frac{1+2\mathrm{cos}⁡2x+{\mathrm{cos}}^{2}⁡2x}{2}+9(1-\mathrm{cos}⁡2x))$$256cos2⁡2x+1−32cos⁡2x=4(21+2cos⁡2x+cos2⁡2x​+9(1−cos⁡2x))

$$256{\mathrm{cos}}^{2}⁡2x+1-32\text{cos}⁡2x=2(19-16\mathrm{cos}⁡2x+{\mathrm{cos}}^{2}⁡2x)$$256cos2⁡2x+1−32cos⁡2x=2(19−16cos⁡2x+cos2⁡2x)

$$\Rightarrow 254{\mathrm{cos}}^{2}⁡2x=37\Rightarrow {\mathrm{cos}}^{2}⁡2x=\frac{37}{254}\Rightarrow \mathrm{cos}⁡2x=\pm \sqrt{\frac{37}{254}}\in [-1, 1]$$⇒254cos2⁡2x=37⇒cos2⁡2x=25437​⇒cos⁡2x=±25437​​∈[−1,1]

Since, the solutions lie in all four quadrants, there are $8$8 solutions which satisfy the equation.