The maximum value of \(\sin\left(x + \dfrac{\pi}{6}\right) + \cos\left(x + \dfrac{\pi}{6}\right)\) is attained at \(x = \)

Let \(f(x) = \sin\left(x + \dfrac{\pi}{6}\right) + \cos\left(x + \dfrac{\pi}{6}\right)\)

Multiplying and dividing by \(\sqrt{2}\):

\(f(x) = \sqrt{2} \left[ \dfrac{1}{\sqrt{2}} \sin\left(x + \dfrac{\pi}{6}\right) + \dfrac{1}{\sqrt{2}} \cos\left(x + \dfrac{\pi}{6}\right) \right]\)

\(f(x) = \sqrt{2} \left[ \cos\dfrac{\pi}{4} \sin\left(x + \dfrac{\pi}{6}\right) + \sin\dfrac{\pi}{4} \cos\left(x + \dfrac{\pi}{6}\right) \right]\)

\(f(x) = \sqrt{2} \sin\left(x + \dfrac{\pi}{6} + \dfrac{\pi}{4}\right)\)

\(f(x) = \sqrt{2} \sin\left(x + \dfrac{5\pi}{12}\right)\)

The maximum value of \(f(x)\) is attained when \(\sin\left(x + \dfrac{5\pi}{12}\right) = 1\)

\(x + \dfrac{5\pi}{12} = \dfrac{\pi}{2}\)

\(x = \dfrac{\pi}{2} - \dfrac{5\pi}{12} = \dfrac{6\pi - 5\pi}{12} = \dfrac{\pi}{12}\)

Answer: \(\dfrac{\pi}{12}\)