Question

For what value of k the equation $\sin x+\cos (k+x)+\cos (k-x)=2$ has real solutions?

• A. $n\pi -\frac{\pi }{3}\le k\le n\pi +\frac{\pi }{3}$, $nϵZ$
• B. $n\pi -\frac{\pi }{4}\le k\le n\pi +\frac{\pi }{4}$, $nϵZ$
• C. $n\pi -\frac{\pi }{12}\le k\le n\pi +\frac{\pi }{12}$, $nϵZ$
• D. $n\pi -\frac{\pi }{6}\le k\le n\pi +\frac{\pi }{6}$, $nϵZ$

Answer 1

The correct option is D $n\pi -\frac{\pi }{6}\le k\le n\pi +\frac{\pi }{6}$, $nϵZ$

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

$\Rightarrow$ $2\cos x\cos k+\sin x=2$

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, $\left|c\right|\le \sqrt{a^2+b^2}$, we have
$2\le \sqrt{1+4{\cos }^{2}k}$

${\cos }^{2}k\ge \frac{3}{4}$

$\Rightarrow$ ${\sin }^{2}k\le \frac{1}{4}$

${\sin }^{2}k-\frac{1}{4}\le 0$

$(\sin k+\frac{1}{2})(\sin k-\frac{1}{2})\le 0$

$-\frac{1}{2}\le \sin k\le \frac{1}{2}$

$\Rightarrow$ $n\pi -\frac{\pi }{6}\le k\le n\pi +\frac{\pi }{6}$, $n\in Z$