Đáp án:
a) \(x = \dfrac{\pi }{6} + k\pi \)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:\sin x \ne 0 \to x \ne k\pi \\
\sqrt 3 \cot x - 3 = 0\\
\to \cot x = \sqrt 3 \\
\to \tan x = \dfrac{1}{{\sqrt 3 }}\\
\to x = \dfrac{\pi }{6} + k\pi \left( {k \in Z} \right)\\
2)\sqrt 3 \cos 2x - \sin 2x - 1 = 0\\
\to \dfrac{{\sqrt 3 }}{2}\cos 2x - \dfrac{1}{2}\sin 2x = \dfrac{1}{2}\\
\to \sin \dfrac{\pi }{3}.\cos 2x - \cos \dfrac{\pi }{3}.\sin 2x = \dfrac{1}{2}\\
\to \sin \left( {\dfrac{\pi }{3} - 2x} \right) = \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
\dfrac{\pi }{3} - 2x = \dfrac{\pi }{6} + k2\pi \\
\dfrac{\pi }{3} - 2x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = - \dfrac{\pi }{4} + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
( Câu 2 t sửa cot2x thành cos2x bài mới làm đc nha )
