Question

If \[ \sin\left(x+\frac{\pi}{3}\right)+\sin\left(x-\frac{\pi}{3}\right)=1, \] then the value of \(x\) in the interval \([0,\pi]\) is

• A. \(\dfrac{\pi}{2}\)
• B. \(\dfrac{\pi}{3}\)
• C. \(0\)
• D. \(\dfrac{\pi}{4}\)

Hint:

Use the identity \(\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}\) when two sine terms differ only by a constant angle.

Answer 1

The Correct Option is A

Step 1: Use sum-to-product on the left side.
\(\sin\left(x+\frac{\pi}{3}\right)+\sin\left(x-\frac{\pi}{3}\right)=2\sin x\cos\frac{\pi}{3}=2\sin x\cdot\frac{1}{2}=\sin x\).

Step 2: Solve sin x = 1 on [0, pi].
\(\sin x=1\Rightarrow x=\dfrac{\pi}{2}\). \[ \boxed{\dfrac{\pi}{2}} \]