JEE Main 2018 — Mathematics Trigonometry

Given equation $8\mathrm{cos}x.(\mathrm{cos}(\frac{\pi }{6}+x).\mathrm{cos}(\frac{\pi }{6}-x)-\frac{1}{2})=1$8cosx.(cos(6π​+x).cos(6π​−x)−21​)=1

We would use the formula

$$\mathrm{cos}(A+B).\mathrm{cos}(A-B)={\mathrm{cos}}^{2}A-{\mathrm{sin}}^{2}B$$cos(A+B).cos(A−B)=cos2A−sin2B

Now, $\mathrm{cos}(\frac{\pi }{6}+x).\mathrm{cos}(\frac{\pi }{6}-x)={\mathrm{cos}}^{2}\frac{\pi }{6}-{\mathrm{sin}}^{2}x$cos(6π​+x).cos(6π​−x)=cos26π​−sin2x

$$=\frac{3}{4}-{\mathrm{sin}}^{2}x$$=43​−sin2x

$$\Rightarrow 8\mathrm{cos}x (\frac{3}{4}-{\mathrm{sin}}^{2}x-\frac{1}{2})=1$$⇒8cosx(43​−sin2x−21​)=1

$$\Rightarrow 8\mathrm{cos}x (\frac{1}{4}-{\mathrm{sin}}^{2}x)=1$$⇒8cosx(41​−sin2x)=1

$$\Rightarrow 8\mathrm{cos}x(\frac{1}{4}-(1-{\mathrm{cos}}^{2}x))=1$$⇒8cosx(41​−(1−cos2x))=1

$$\Rightarrow 8\mathrm{cos}x({\mathrm{cos}}^{2}x-\frac{3}{4})=1$$⇒8cosx(cos2x−43​)=1

$$\Rightarrow 8{\mathrm{cos}}^{3}x-6\mathrm{cos}x=1$$⇒8cos3x−6cosx=1

$$\Rightarrow 2(4{\mathrm{cos}}^{3}x-3\mathrm{cos}x)=1$$⇒2(4cos3x−3cosx)=1

$$\Rightarrow 2\mathrm{cos}3x=1$$⇒2cos3x=1

$$\Rightarrow \mathrm{cos}3x=\frac{1}{2}$$⇒cos3x=21​

General Solution $3x=2n\pi \pm \frac{\pi }{3}$3x=2nπ±3π​

$$\Rightarrow x=\frac{2n\pi }{3}\pm \frac{\pi }{9}$$⇒x=32nπ​±9π​

$$x\in [0, \pi ]$$x∈[0,π]

$$\therefore x=\frac{\pi }{9},\frac{2\pi }{3}-\frac{\pi }{9}, \frac{2\pi }{3}+\frac{\pi }{9}$$∴x=9π​,32π​−9π​,32π​+9π​

$$\mathrm{Sum}=\frac{13\pi }{9}, k=\frac{13}{9}$$Sum=913π​,k=913​