Question

If the equation $\sin x+\cos (k+x)+\cos (k-x)=2$ has real solutions, then $\sin k$ can be

• A. $\frac{1}{4}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{3}{4}$
• D. $\frac{-3}{4}$

Answer 1

Answer: (A)

$\sin x+2\cos k\cdot \cos x=2$
$\therefore 2\le \sqrt{4{\cos }^{2}k+1}$
$4\le 4{\cos }^{2}k+1$
$3\le 4-4{\sin }^{2}k$
$4{\sin }^{2}k\le 1$
$\sin k\in \left[\frac{-1}{2},\frac{1}{2}\right]$