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 the identity for sum of sines.
We know that \[ \sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \] Here, \[ A=x+\frac{\pi}{3} \] and \[ B=x-\frac{\pi}{3} \]

Step 2: Substitute in the identity.
\[ \sin\left(x+\frac{\pi}{3}\right)+\sin\left(x-\frac{\pi}{3}\right) \] \[ = 2\sin\left(\frac{x+\frac{\pi}{3}+x-\frac{\pi}{3}}{2}\right) \cos\left(\frac{x+\frac{\pi}{3}-x+\frac{\pi}{3}}{2}\right) \] \[ =2\sin x\cos\frac{\pi}{3} \] Since \[ \cos\frac{\pi}{3}=\frac{1}{2}, \] we get \[ 2\sin x\cdot \frac{1}{2}=\sin x \]

Step 3: Use the given equation.
Given, \[ \sin\left(x+\frac{\pi}{3}\right)+\sin\left(x-\frac{\pi}{3}\right)=1 \] So, \[ \sin x=1 \]

Step 4: Find \(x\) in \([0,\pi]\).
In the interval \[ [0,\pi], \] we have \[ \sin x=1 \] at \[ x=\frac{\pi}{2}. \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{\pi}{2}} \]