Mathematics - Binomial Theorem Question with Solution | TestHub

We have, $8{\tan }^2\frac{x}{2}=1+\sec x$ $\begin{array}{rl}& \Rightarrow 8\left(\frac{1-\cos x}{1+\cos x}\right)=1+\frac{1}{\cos x}=\frac{1+\cos x}{\cos x}\\ & \Rightarrow 8\cos x-8{\cos }^{2}x=(1+\cos x{)}^2\\ & \Rightarrow 9{\cos }^2\mathrm{x}-6\cos \mathrm{x}+1=0\\ & \Rightarrow (3\cos \mathrm{x}-1{)}^2=0\Rightarrow 3\cos \mathrm{x}-1=0\\ & \Rightarrow \cos \mathrm{x}=\frac{1}{3}=\cos \alpha \text{ (say) }\Rightarrow \mathrm{x}=2\mathrm{n}\pi \pm \alpha \\ & \therefore \mathrm{x}=2\mathrm{n}\pi \pm {\cos }^{-1}\left(\frac{1}{3}\right)\text{ where }\mathrm{n}\in \mathrm{I}\end{array}$