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

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

Complex Numbers and Quadratic Equations

Solution:

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