Question

If $\cos 2x=(\sqrt{2}+1)(\cos x-(\frac{1}{\sqrt{2}}\left)\right)$ then $\cos x\ne \frac{1}{\sqrt{2}},x\in$

• A. $\left\{2n\pi \pm \frac{\pi }{3}:n\in Z\right\}$
• B. $\left\{2n\pi \pm \frac{\pi }{6}:n\in Z\right\}$
• C. $\left\{2n\pi \pm \frac{\pi }{2}:n\in Z\right\}$
• D. $\left\{2n\pi \pm \frac{\pi }{4}:n\in Z\right\}$

Answer 1

The correct option is A

$\left\{2n\pi \pm \frac{\pi }{3}:n\in Z\right\}$

Explanation for the correct option:

Given,

$\cos 2x=(\sqrt{2}+1)(\cos x-(\frac{1}{\sqrt{2}}\left)\right)$

$\Rightarrow$ $(2{\cos }^{2}x–1)=\left[\frac{(\sqrt{2}+1)}{\sqrt{2}}\right](\sqrt{2}\cos x–1)$

$\Rightarrow$$(\sqrt{2}\cos x+1)(\sqrt{2}\cos x–1)=\left[\frac{(\sqrt{2}+1)}{\sqrt{2}}\right](\sqrt{2}\cos x–1)$

$\Rightarrow$ $\sqrt{2}\cos x+1=\left[\frac{(\sqrt{2}+1)}{\sqrt{2}}\right]$

$\Rightarrow$ $\sqrt{2}\cos x=1+\left(\frac{1}{\sqrt{2}}\right)–1$

$\Rightarrow$ $\cos x=\frac{1}{\left(\sqrt{2}\right)\left(\sqrt{2}\right)}$

$\Rightarrow$ $\cos x=\frac{1}{2}$

$\mathbf{\therefore }\mathit{x}\mathbf{\in }\mathbf{}\mathbf{2}\mathit{n}\mathit{\pi }\mathbf{}\mathbf{\pm }\mathbf{}\frac{\mathbf{\pi }}{\mathbf{3}}\mathbf{:}\mathbf{}\mathit{n}\mathbf{}\mathbf{\in }\mathbf{}\mathit{Z}$

Hence, the correct option is (A).