Question

$\text{If}\cos x+\cos (k+x)-\cos (k-x)=2$ has real solutions, then

• A. $k\in \left[n\pi +\frac{\pi }{3},n\pi +\frac{2\pi }{3}\right]$
• B. $k\in \left[2n\pi +\frac{\pi }{3},2n\pi -\frac{2\pi }{3}\right]$
• C. $k\in \left[n\pi -\frac{\pi }{6},n\pi +\frac{\pi }{6}\right]$
• D. $k\in \left[2n\pi -\frac{\pi }{6},2n\pi +\frac{\pi }{6}\right]$

Answer 1

The correct option is A $k\in \left[n\pi +\frac{\pi }{3},n\pi +\frac{2\pi }{3}\right]$

$\cos x+\cos (k+x)-\cos (k-x)=2\Rightarrow \cos x-2\sin k\cdot \sin x=2$

We know that for an equation $a\cos x+b\sin x=c$ to have solution $|c|\le \sqrt{a^2+b^2}$

$\sqrt{1+4{\sin }^{2}k}\ge 2\Rightarrow 1+4{\sin }^{2}k\ge 4\Rightarrow |\sin k|\ge \frac{\sqrt{3}}{2}\Rightarrow \sin k\ge \frac{\sqrt{3}}{2}\text{or}\sin k\le -\frac{\sqrt{3}}{2}\Rightarrow k\in [2n\pi +\frac{\pi }{3},2n\pi +\frac{2\pi }{3}]\text{or}k\in \left[\right(2n+1)\pi +\frac{\pi }{3},(2n+1)\pi +\frac{2\pi }{3}]\Rightarrow k\in \left[n\pi +\frac{\pi }{3},n\pi +\frac{2\pi }{3}\right]$