Question

The solution of inequality $\cos 2 x \leq \cos x$ is

• A. $x \in\left[2 n \pi-\frac{\pi}{3}, 2 n \pi+\frac{\pi}{3}\right], n \in I$
• B. $x \in\left[2 n \pi-\frac{2 \pi}{3}, 2 n \pi+\frac{2 \pi}{3}\right], n \in I$
• C. $x \in\left[2 n \pi, 2 n \pi+\frac{2 \pi}{3}\right], n \in I$
• D. $x \in\left[2 n \pi-\frac{2 \pi}{3}, 2 n \pi\right], n \in I$

Answer 1

Answer: (B)

$\cos 2 x \leq \cos x \Rightarrow 2 \cos ^2 x-\cos x-1 \leq 0$
$ \Rightarrow 2 \cos ^2 x-2 \cos x+\cos x-1 \leq 0 $
$\Rightarrow 2 \cos x(\cos x-1)+1(\cos x-1) \leq 0$
$ \Rightarrow(\cos x-1)(2 \cos x+1) \leq 0 $
$\Rightarrow \cos x \in\left[-\frac{1}{2}, 1\right]$
$ \therefore x \in\left[-\frac{2 \pi}{3}, \frac{2 \pi}{3}\right]$
General soltution is $x \in\left[2 n \pi-\frac{2 \pi}{3}, 2 n \pi+\frac{2 \pi}{3}\right], n \in I$