Ngân hàng bài tập
B
Giải phương trình $$\tan^23x-\left(\sqrt{3}-1\right)\tan3x-\sqrt{3}=0$$
 | \(\left[\begin{array}{l}x=-\dfrac{\pi}{12}+k\dfrac{\pi}{3}\\ x=\dfrac{\pi}{9}+k\dfrac{\pi}{3}\end{array}\right.\,(k\in\mathbb{Z})\) |
 | \(x=-\dfrac{\pi}{12}+k\dfrac{\pi}{3}\,(k\in\mathbb{Z})\) |
 | \(\left[\begin{array}{l}x=-\dfrac{\pi}{12}+k\dfrac{2\pi}{3}\\ x=\dfrac{2\pi}{9}+k\dfrac{2\pi}{3}\end{array}\right.\,(k\in\mathbb{Z})\) |
 | \(\left[\begin{array}{l}x=-\dfrac{\pi}{12}+k\dfrac{\pi}{3}\\ x=\dfrac{\pi}{18}+ k\dfrac{\pi}{3}\end{array}\right.\,(k\in\mathbb{Z})\) |
1 lời giải
Chọn phương án A.
\(\begin{aligned}
&\tan^23x-\left(\sqrt{3}-1\right)\tan3x-\sqrt{3}=0\\
\Leftrightarrow&\left[\begin{array}{l}\tan3x=\sqrt{3}\\ \tan3x=-1\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}3x=\dfrac{\pi}{3}+k\pi\\ 3x=-\dfrac{\pi}{4}+k\pi\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}x=\dfrac{\pi}{9}+k\dfrac{\pi}{3}\\ x=-\dfrac{\pi}{12}+k\dfrac{\pi}{3}.\end{array}\right.\,(k\in\mathbb{Z})
\end{aligned}\)