Ngân hàng bài tập
B
Nghiệm của phương trình \(\sin x+\sqrt{3}\cos x=1\) là
 | \(x=\dfrac{-\pi}{6}+k2\pi(k\in\mathbb{Z})\) |
 | \(\left[\begin{array}{l}x=\dfrac{-\pi}{6}+k2\pi\\ x=\dfrac{\pi}{2}+k2\pi\end{array}\right.\,(k\in\mathbb{Z})\) |
 | \(\left[\begin{array}{l}x=\dfrac{-\pi}{6}+k\pi\\ x=\dfrac{\pi}{2}+k\pi\end{array}\right.\,(k\in\mathbb{Z})\) |
 | \(\left[\begin{array}{l}x=k2\pi\\ x=\dfrac{\pi}{3}+k2\pi\end{array}\right.\,(k\in\mathbb{Z})\) |
1 lời giải
Chọn phương án B.
\(\begin{eqnarray*}
&\sin x+\sqrt{3}\cos x&=1\\
\Leftrightarrow&\dfrac{1}{2}\sin x+\dfrac{\sqrt{3}}{2}\cos x&=\dfrac{1}{2}\\
\Leftrightarrow&\cos\dfrac{\pi}{3}\sin x+\sin\dfrac{\pi}{3}\cos x&=\dfrac{1}{2}\\
\Leftrightarrow&\sin\left(x+\dfrac{\pi}{3}\right)&=\dfrac{1}{2}\\
\Leftrightarrow&\left[\begin{array}{l}x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\ x+\dfrac{\pi}{3}=\dfrac{5\pi}{6}+k2\pi\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}x=\dfrac{-\pi}{6}+k2\pi\\ x=\dfrac{\pi}{2}+k2\pi\end{array}\right.&(k\in\mathbb{Z})
\end{eqnarray*}\)