Question

The solution of the equation

$\mathbf{\cos}^{\mathbf{2}}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{\cos}\mathbf{x}\mathbf{= 4}\mathbf{\sin}\mathbf{x}\mathbf{-}\mathbf{\sin}\mathbf{2}\mathbf{x,(0}\mathbf{\leq}\mathbf{x}\mathbf{\leq}\mathbf{\pi)}$ is

• A. $\pi - \cot^{- 1}\left( \frac{1}{2} \right)$
• B. $\pi - \tan^{- 1}(2)$
• C. $\pi + \tan^{- 1}\left( - \frac{1}{2} \right)$
• D. None of these

Answer 1

Answer: (C)

$\pi + \tan^{- 1}\left( - \frac{1}{2} \right)$

Answer 2

Given equation is $\cos^{2}x - 2\cos x = 4\sin x - \sin 2x$

$\mathbf{\Rightarrow}\mathbf{\cos}^{\mathbf{2}}\mathbf{x}\mathbf{- 2}\mathbf{\cos}\mathbf{x}\mathbf{= 4}\mathbf{\sin}\mathbf{x}\mathbf{- 2}\mathbf{\sin}\mathbf{x}\mathbf{\cos}\mathbf{x}\mathbf{\Rightarrow}\mathbf{\cos}\mathbf{x}\mathbf{(}\mathbf{\cos}\mathbf{x}\mathbf{- 2) = 2}\mathbf{\sin}\mathbf{x}\mathbf{(2 -}\mathbf{\cos}\mathbf{x}\mathbf{)}\mathbf{\Rightarrow (}\mathbf{\cos}\mathbf{x}\mathbf{- 2)(}\mathbf{\cos}\mathbf{x}\mathbf{+ 2}\mathbf{\sin}\mathbf{x}\mathbf{) = 0}\mathbf{\Rightarrow}\mathbf{\cos}\mathbf{x}\mathbf{+ 2}\mathbf{\sin}\mathbf{x}\mathbf{= 0(\because}\mathbf{\cos}\mathbf{x}\mathbf{\neq 2) \Rightarrow}\mathbf{\tan}\mathbf{x}\mathbf{= -}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\Rightarrow}\mathbf{x = n\pi +}\mathbf{\tan}^{\mathbf{-}\mathbf{1}}\left( \mathbf{-}\frac{\mathbf{1}}{\mathbf{2}} \right)\mathbf{,n}\mathbf{\in}\mathbf{I}$As

$0 \leq x \leq \pi,$ therefore $x = \pi + \tan^{- 1}\left( - \frac{1}{2} \right)$.