Question

Find the value(s) of $k$ for which the equation $\sin x+\cos (k+x)+\cos (k-x)=2$ has real solutions.

• A. $n\pi -\frac{\pi }{3}\le k\le n\pi +\frac{\pi }{3},n\in I$
• B. $n\pi -\frac{2\pi }{3}\le k\le n\pi +\frac{2\pi }{3},n\in I$
• C. $n\pi -\frac{\pi }{6}\le k\le n\pi +\frac{\pi }{6},n\in I$
• D. $n\pi -\frac{3\pi }{4}\le k\le n\pi +\frac{3\pi }{4},n\in I$

Answer 1

The correct option is C $n\pi -\frac{\pi }{6}\le k\le n\pi +\frac{\pi }{6},n\in I$

Given, $\sin x+\cos (k+x)+\cos (k-x)=2$
This equation is of the form $a\cos x+b\sin x=c$
Here $a=2\cos k,b=1$ and $c=2$
Since for real solutions, $|c|\le \sqrt{a^2+b^2}$
$\therefore |2|\le \sqrt{1+4{\cos }^{2}k}\Rightarrow 2\le \sqrt{1+4{\cos }^{2}k}$
$\Rightarrow {\cos }^{2}k\ge \frac{3}{4}\Rightarrow {\sin }^{2}k\le \frac{1}{4}$
$\Rightarrow {\sin }^{2}k=\frac{1}{4}\le 0\Rightarrow [\sin k+\frac{1}{2}][\sin k-\frac{1}{2}]\le 0$
$\Rightarrow -\frac{1}{2}\le \sin k\le \frac{1}{2}\Rightarrow n\pi -\frac{\pi }{6}\le k\le n\pi +\frac{\pi }{6}$