Ngân hàng bài tập
B
[Đề kiểm tra 1 tiết Đại Số & Giải Tích chương 1 lớp 11 Trường THPT Tân Lược - Vĩnh Long (2019-2020)]
Giải phương trình $$4\sin x\cdot\cos3x=1-2\sin2x$$
 | \(\left[\begin{array}{l}x=\dfrac{\pi}{6}+k2\pi\\ x=\dfrac{5\pi}{6}+k2\pi\end{array}\right.\,\left(k\in\mathbb{Z}\right)\) |
 | \(\left[\begin{array}{l}x=\dfrac{\pi}{6}+k\pi\\ x=\dfrac{5\pi}{6}+k\pi\end{array}\right.\,\left(k\in\mathbb{Z}\right)\) |
 | \(\left[\begin{array}{l}x=\dfrac{\pi}{24}+\dfrac{k\pi}{2}\\ x=\dfrac{5\pi}{24}+\dfrac{k\pi}{2}\end{array}\right.\,\left(k\in\mathbb{Z}\right)\) |
 | \(\left[\begin{array}{l}x=\dfrac{\pi}{24}+k2\pi\\ x=\dfrac{5\pi}{24}+k2\pi\end{array}\right.\,\left(k\in\mathbb{Z}\right)\) |
1 lời giải
Chọn phương án C.
\(\begin{eqnarray*}
&4\sin x\cdot\cos3x&=1-2\sin2x\\
\Leftrightarrow&4\cdot\dfrac{1}{2}\left[\sin(x+3x)+\sin(x-3x)\right]&=1-2\sin2x\\
\Leftrightarrow&2\left[\sin4x+\sin(-2x)\right]&=1-2\sin2x\\
\Leftrightarrow&2\left[\sin4x-\sin2x\right]&=1-2\sin2x\\
\Leftrightarrow&2\sin4x-2\sin2x&=1-2\sin2x\\
\Leftrightarrow&2\sin4x&=1\\
\Leftrightarrow&\sin4x&=\dfrac{1}{2}\\
\Leftrightarrow&\left[\begin{array}{l}4x=\dfrac{\pi}{6}+k2\pi\\ 4x=\dfrac{5\pi}{6}+k2\pi\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}x=\dfrac{\pi}{24}+\dfrac{k\pi}{2}\\ x=\dfrac{5\pi}{24}+\dfrac{k\pi}{2}\end{array}\right.\,\left(k\in\mathbb{Z}\right)
\end{eqnarray*}\)