Question

If $\cos x-\sin x\ge 1$ and $0\le x\le 2\pi$, then the solution set for $x$ is

• A. $[0,\frac{\pi }{4}]\cup [\frac{7\pi }{4},2\pi ]$
• B. $[\frac{3\pi }{2},\frac{7\pi }{4}]\cup \left\{0\right\}$
• C. $[\frac{7\pi }{4},2\pi ]\cup \left\{0\right\}$
• D. none of these

Answer 1

The correct option is C $[\frac{7\pi }{4},2\pi ]\cup \left\{0\right\}$

$\cos x-\sin x\ge 1$

$\Rightarrow \frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x\ge \frac{1}{\sqrt{2}}$

$\Rightarrow \cos \left(\frac{\pi }{4}\right).cosx-\sin \left(\frac{\pi }{4}\right).\sin x\ge \frac{1}{\sqrt{2}}$

$\Rightarrow \cos (\frac{\pi }{4}+x)\ge \frac{1}{\sqrt{2}}$

$\Rightarrow \frac{1}{\sqrt{2}}\le \cos (\frac{\pi }{4}+x)\le 1$
Hence solution in given interval is,

$x\in [\frac{7\pi }{4},2\pi ]\cup 0$
Hence, option 'C' is correct.