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Ifsin $\left(x+\frac{\pi}{3}\right)+\sin \left(x-\frac{\pi}{3}\right)=1$, then the value of $x$ in the interval $[0, \pi]$
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The correct answer is:
$\frac{\pi}{2}$
$\sin \left(x+\frac{\pi}{3}\right)+\sin \left(x-\frac{\pi}{3}\right)=1$
$\Rightarrow \quad 2 \sin x \cdot \cos \frac{\pi}{3}=1$
$[\because 2 \sin A \cdot \cos B=\sin (A+B)+\sin (A-B)]$
$\Rightarrow \quad 2 \sin x \cdot \frac{1}{2}=1$
$\sin x=1 \Rightarrow x=\frac{\pi}{2} \quad[\because x \in[0, \pi]]$
$\Rightarrow \quad 2 \sin x \cdot \cos \frac{\pi}{3}=1$
$[\because 2 \sin A \cdot \cos B=\sin (A+B)+\sin (A-B)]$
$\Rightarrow \quad 2 \sin x \cdot \frac{1}{2}=1$
$\sin x=1 \Rightarrow x=\frac{\pi}{2} \quad[\because x \in[0, \pi]]$
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