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Câu hỏi: Phương trình $\sqrt{3}\text{sin}x+\text{cos}x=-1$ tương đương với phương trình nào sau đây?
A. $\text{sin}\left( x+\dfrac{\pi }{6} \right)=-\dfrac{1}{2}$.
B. $\text{sin}\left( x+\dfrac{\pi }{6} \right)=\dfrac{1}{2}$.
C. $\text{sin}\left( x-\dfrac{\pi }{6} \right)=-\dfrac{1}{2}$.
D. $\text{sin}\left( x-\dfrac{\pi }{6} \right)=\dfrac{1}{2}$.
A. $\text{sin}\left( x+\dfrac{\pi }{6} \right)=-\dfrac{1}{2}$.
B. $\text{sin}\left( x+\dfrac{\pi }{6} \right)=\dfrac{1}{2}$.
C. $\text{sin}\left( x-\dfrac{\pi }{6} \right)=-\dfrac{1}{2}$.
D. $\text{sin}\left( x-\dfrac{\pi }{6} \right)=\dfrac{1}{2}$.
Phương pháp:
$\sin a\cos b+\cos a\sin b=\sin \left( a+b \right)$
Cách giải:
Ta có: $\sqrt{3}\sin x+\cos x=-1\Leftrightarrow \dfrac{\sqrt{3}}{2}\sin x+\dfrac{1}{2}\cos x=-\dfrac{1}{2}\Leftrightarrow \cos \dfrac{\pi }{6}\sin x+\sin \dfrac{\pi }{6}\cos x=-\dfrac{1}{2}\Leftrightarrow \sin \left( x+\dfrac{\pi }{6} \right)=-\dfrac{1}{2}.$
$\sin a\cos b+\cos a\sin b=\sin \left( a+b \right)$
Cách giải:
Ta có: $\sqrt{3}\sin x+\cos x=-1\Leftrightarrow \dfrac{\sqrt{3}}{2}\sin x+\dfrac{1}{2}\cos x=-\dfrac{1}{2}\Leftrightarrow \cos \dfrac{\pi }{6}\sin x+\sin \dfrac{\pi }{6}\cos x=-\dfrac{1}{2}\Leftrightarrow \sin \left( x+\dfrac{\pi }{6} \right)=-\dfrac{1}{2}.$
Đáp án A.