Ngân hàng bài tập
B
Giải phương trình \(2\sin\left(4x-\dfrac{\pi}{3}\right)-1=0\).
 | \(\left[\begin{array}{l}x=\pi+k2\pi\\ x=k\dfrac{\pi}{2}\end{array}\right.\,(k\in\mathbb{Z})\) |
 | \(\left[\begin{array}{l}x=k\pi\\ x=\pi+k2\pi\end{array}\right.\,(k\in\mathbb{Z})\) |
 | \(\left[\begin{array}{l}x=k2\pi\\ x=\dfrac{\pi}{2}+k2\pi\end{array}\right.\,(k\in\mathbb{Z})\) |
 | \(\left[\begin{array}{l}x=\dfrac{\pi}{8}+k\dfrac{\pi}{2}\\ x=\dfrac{7\pi}{24}+k\dfrac{\pi}{2}\end{array}\right.\,(k\in\mathbb{Z})\) |
1 lời giải
Chọn phương án D.
\(\begin{aligned}
2\sin\left(4x-\dfrac{\pi}{3}\right)-1=0\Leftrightarrow&\sin\left(4x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\\
\Leftrightarrow&\left[\begin{array}{l}4x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\ 4x-\dfrac{\pi}{3}=\dfrac{5\pi}{6}+k2\pi\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}4x=\dfrac{\pi}{2}+k2\pi\\ 4x=\dfrac{7\pi}{6}+k2\pi\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}x=\dfrac{\pi}{8}+k\dfrac{\pi}{2}\\ x=\dfrac{7\pi}{24}+k\dfrac{\pi}{2}\end{array}\right.\,(k\in\mathbb{Z})
\end{aligned}\)