Question

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

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

Answer 1

Answer: (B)

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