Đáp án:
min = -1
Giải thích các bước giải:
\(\begin{array}{l}
\,\,\,\,\sin x + \sin \left( {x + \dfrac{{2\pi }}{3}} \right)\\
= \sin x + \sin x\left( { - \dfrac{1}{2}} \right) + \cos x.\dfrac{{\sqrt 3 }}{2}\\
= \dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x\\
= \sin x\cos \dfrac{\pi }{3} + \cos x\sin \dfrac{\pi }{3}\\
= \sin \left( {x + \dfrac{\pi }{3}} \right)\\
- 1 \le \sin \left( {x + \dfrac{\pi }{3}} \right) \le 1\\
\Rightarrow - 1 \le \,\sin x + \sin \left( {x + \dfrac{{2\pi }}{3}} \right) \le 1\\
\Rightarrow \min = - 1
\end{array}\)
