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The maximum value of $\sin \left(x+\frac{\pi}{6}\right)+\cos \left(x+\frac{\pi}{6}\right)$ in the interval $\left(0, \frac{\pi}{2}\right)$ is attained at
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The correct answer is:
$x=\frac{\pi}{12}$
$\sqrt{2} \cos \left(x+\frac{\pi}{6}-\frac{\pi}{4}\right)=\sqrt{2} \cos \left(x-\frac{\pi}{12}\right)$
Hence maximum value will be at $x=\frac{\pi}{12}$.
Hence maximum value will be at $x=\frac{\pi}{12}$.
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