Q68 / 160 · AP EAMCET 2005 · Mathematics
If $\quad \cos 2 x=(\sqrt{2}+1)\left(\cos x-\frac{1}{\sqrt{2}}\right), \cos x \neq \frac{1}{2}$, then $x \in$
- A$\left\{2 n \pi \pm \frac{\pi}{3}: n \in Z\right\}$
- B$\left\{2 n \pi \pm \frac{\pi}{6}: n \in Z\right\}$
- C$\left\{2 n \pi \pm \frac{\pi}{2}: n \in Z\right\}$
- Correct answerD$\left\{2 n \pi \pm \frac{\pi}{4}: n \in Z\right\}$
Solution
$\begin{aligned}
& \cos 2 x=(\sqrt{2}+1)\left(\cos x-\frac{1}{\sqrt{2}}\right) \\
& \Rightarrow \quad \cos 2 x=\sqrt{2} \cos x-1+\cos x-\frac{1}{\sqrt{2}} \\
& \Rightarrow 1+\cos 2 x=\cos x(\sqrt{2}+1)-\frac{1}{\sqrt{2}} \\
& \Rightarrow 2 \cos ^2 x-\cos x(\sqrt{2}+1)+\frac{1}{\sqrt{2}}=0 \\
& \Rightarrow \quad \cos x=\frac{(\sqrt{2}+1) \pm \sqrt{(\sqrt{2}+1)^2-\frac{8}{\sqrt{2}}}}{2(2)} \\
& =\frac{(\sqrt{2}+1) \pm \sqrt{3+2 \sqrt{2}-4 \sqrt{2}}}{4} \\
& =\frac{(\sqrt{2}+1) \pm \sqrt{(\sqrt{2}-1)^2}}{4} \\
& =\frac{\sqrt{2}+1 \pm(\sqrt{2}-1)}{4} \\
&
\end{aligned}$
on taking + ve sign
$\begin{aligned}
\cos x & =\frac{\sqrt{2}+1+\sqrt{2}-1}{4}=\frac{2 \sqrt{2}}{4}=\frac{1}{\sqrt{2}} \\
\Rightarrow \quad x & =2 n \pi \pm \frac{\pi}{4} \forall n \in Z
\end{aligned}$
Hence, the correct answer is option: 'D'.