JEE Main 2019 — Mathematics Trigonometry

We have, ${cos}^{-1}\frac{2}{3x}+{cos}^{-1}\frac{3}{4x}=\frac{\pi }{2},x>\frac{3}{4}...(i)$cos−13x2​+cos−14x3​=2π​,x>43​...(i)$\therefore {\mathrm{cos}}^{-1}\frac{3}{4x}={\mathrm{sin}}^{-1}\frac{\sqrt{16{x}^{2}-9}}{4x}$∴cos−14x3​=sin−14x16x2−9​​

Put in equation $(i)$(i) we get,

$$\Rightarrow {cos}^{-1}\frac{2}{3x}+{sin}^{-1}\frac{\sqrt{16{x}^{2}-9}}{4x}=\frac{\pi }{2}...(\mathrm{ii})$$⇒cos−13x2​+sin−14x16x2−9​​=2π​...(ii)

We know ${\mathrm{sin}}^{-1}x+{\mathrm{cos}}^{-1}x=\frac{\pi }{2}$sin−1x+cos−1x=2π​

From equation $(\mathrm{ii})$(ii) we get,

$$\Rightarrow \frac{2}{3x}=\frac{\sqrt{16{x}^{2}-9}}{4x}$$⇒3x2​=4x16x2−9​​

$$\Rightarrow {(\frac{8}{3})}^{2}=16{x}^{2}-9(\because x\neq 0)$$⇒(38​)2=16x2−9(∵x=0)

$$\Rightarrow 16{x}^{2}=\frac{64}{9}+9=\frac{145}{9}$$⇒16x2=964​+9=9145​

$$\Rightarrow {x}^{2}=\frac{145}{144}$$⇒x2=144145​

$$\Rightarrow x=\frac{\sqrt{145}}{12}$$⇒x=12145​​