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Question: Answered & Verified by Expert
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
MathematicsTrigonometric EquationsJEE Main
Options:
  • A $x=\frac{\pi}{12}$
  • B $x=\frac{\pi}{6}$
  • C $x=\frac{\pi}{3}$
  • D $x=\frac{\pi}{2}$
Solution:
2139 Upvotes Verified Answer
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 $x=\frac{\pi}{12}$.

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