JEE Main 2025 — Mathematics Trigonometry

$\begin{aligned} & \cos ^{-1} x=\pi+\sin ^{-1} x+\sin ^{-1}(2 x+1) \\ & 2 \cos ^{-1} x-\sin ^{-1}(2 x+1)=\frac{3 \pi}{2} \\ & 2 \alpha-\beta=\frac{3 \pi}{2} \text { where } \cos ^{-1} x=\alpha, \sin ^{-1}(2 x+1)=\beta \\ & 2 \alpha=\frac{3 \pi}{2}+\beta \\ & \cos 2 \alpha=\sin \beta \\ & 2 \cos ^2 \alpha-1=\sin \beta \\ & 2 x^2-1=2 x+1 \\ & x^2-x-1=0\end{aligned}$​cos−1x=π+sin−1x+sin−1(2x+1)2cos−1x−sin−1(2x+1)=23π​2α−β=23π​ where cos−1x=α,sin−1(2x+1)=β2α=23π​+βcos2α=sinβ2cos2α−1=sinβ2x2−1=2x+1x2−x−1=0​ $\begin{aligned} & \Rightarrow \mathrm{n}=\frac{1 \pm \sqrt{5}}{2}=\left[\begin{array}{c}\mathrm{n}=\frac{1+\sqrt{5}}{2} \text { rejected } \\ \mathrm{n}=\frac{1-\sqrt{5}}{2}\end{array}\right. \\ & \therefore 4 \mathrm{x}^2-4 \mathrm{x}=4 \\ & (2 \mathrm{x}-1)^2=5\end{aligned}$​⇒n=21±5​​=[n=21+5​​ rejected n=21−5​​​∴4x2−4x=4(2x−1)2=5​