Ngân hàng bài tập
B
Giải phương trình $$\left(2\cos\dfrac{x}{2}-1\right)\left(\sin\dfrac{x}{2}+2\right)=0$$
 | \(x=\pm\dfrac{2\pi}{3}+k2\pi,\,k\in\mathbb{Z}\) |
 | \(x=\pm\dfrac{\pi}{3}+k2\pi,\,k\in\mathbb{Z}\) |
 | \(x=\pm\dfrac{\pi}{3}+k4\pi,\,k\in\mathbb{Z}\) |
 | \(x=\pm\dfrac{2\pi}{3}+k4\pi,\,k\in\mathbb{Z}\) |
1 lời giải
Chọn phương án D.
\(\begin{aligned}
&\left(2\cos\dfrac{x}{2}-1\right)\left(\sin\dfrac{x}{2}+2\right)=0\\
\Leftrightarrow&\left[\begin{array}{l}
2\cos\dfrac{x}{2}-1=0\\
\sin\dfrac{x}{2}+2=0
\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{ll}
\cos\dfrac{x}{2}=\dfrac{1}{2}\\
\sin\dfrac{x}{2}=-2 &\text{(vô nghiệm)}
\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}
\dfrac{x}{2}=\dfrac{\pi}{3}+k2\pi\\
\dfrac{x}{2}=-\dfrac{\pi}{3}+k2\pi
\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}
x=\dfrac{2\pi}{3}+k4\pi\\
x=-\dfrac{2\pi}{3}+k4\pi
\end{array}\right.\,k\in\mathbb{Z}
\end{aligned}\)