Đề bài

Giải phương trình: a) \(\sin 3x = \frac{{\sqrt 3 }}{2}\)                                          b) \(\sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right) =  - \frac{{\sqrt 2 }}{2}\) c) \(\cos \left( {3x + \frac{\pi }{3}} \right) =  - \frac{1}{2}\)                                d) \(2\cos x + \sqrt 3  = 0\) e) \(\sqrt 3 \tan x - 1 = 0\)                                    g) \(\cot \left( {x + \frac{\pi }{5}} \right) = 1\)

Lời giải chi tiết

a) Ta có \(\sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2}\), phương trình trở thành:\(\sin 3x = \sin \frac{\pi }{3} \Leftrightarrow \left[ \begin{array}{l}3x = \frac{\pi }{3} + k2\pi \\3x = \pi  - \frac{\pi }{3} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{9} + k\frac{{2\pi }}{3}\\x = \frac{{2\pi }}{9} + k\frac{{2\pi }}{3}\end{array} \right.\)\(\left( {k \in \mathbb{Z}} \right)\)b) Ta có \(\sin \left( { - \frac{\pi }{4}} \right) =  - \frac{{\sqrt 2 }}{2}\), phương trình trở thành:\(\sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right) = \sin \left( { - \frac{\pi }{4}} \right) \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} + \frac{\pi }{4} =  - \frac{\pi }{4} + k2\pi \\\frac{x}{2} + \frac{\pi }{4} = \pi  + \frac{\pi }{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} =  - \frac{\pi }{2} + k2\pi \\\frac{x}{2} = \pi  + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  - \pi  + k4\pi \\x = 2\pi  + k4\pi \end{array} \right.\)\(\left( {k \in \mathbb{Z}} \right)\) c) Ta có \(\cos \frac{{2\pi }}{3} = \frac{{ - 1}}{2}\), phương trình trở thành:\(\cos \left( {3x + \frac{\pi }{3}} \right) = \cos \frac{{2\pi }}{3} \Leftrightarrow \left[ \begin{array}{l}3x + \frac{\pi }{3} = \frac{{2\pi }}{3} + k2\pi \\3x + \frac{\pi }{3} =  - \frac{{2\pi }}{3} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}3x = \frac{\pi }{3} + k2\pi \\3x =  - \pi  + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{9} + k\frac{{2\pi }}{3}\\x =  - \frac{\pi }{3} + k\frac{{2\pi }}{3}\end{array} \right.\)\(\left( {k \in \mathbb{Z}} \right)\)d) \(2\cos x + \sqrt 3  = 0 \Leftrightarrow \cos x =  - \frac{{\sqrt 3 }}{2}\).Ta có: \(\cos \frac{{5\pi }}{6} =  - \frac{{\sqrt 3 }}{2}\), phương trình trở thành: \(\cos x = \cos \frac{{5\pi }}{6} \Leftrightarrow \left[ \begin{array}{l}x = \frac{{5\pi }}{6} + k2\pi \\x =  - \frac{{5\pi }}{6} + k2\pi \end{array} \right.\)\(\left( {k \in \mathbb{Z}} \right)\)e) \(\sqrt 3 \tan x - 1 = 0 \Leftrightarrow \tan x = \frac{1}{{\sqrt 3 }}\)Ta có \(\tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }}\), phương trình trở thành: \(\tan x = \tan \frac{\pi }{6} \Leftrightarrow x = \frac{\pi }{6} + k\pi \)\(\left( {k \in \mathbb{Z}} \right)\)f) Ta có \(\cot \frac{\pi }{4} = 1\), phương trình trở thành:\(\cot \left( {x + \frac{\pi }{5}} \right) = \cot \frac{\pi }{4} \Leftrightarrow x + \frac{\pi }{5} = \frac{\pi }{4} + k\pi  \Leftrightarrow x = \frac{\pi }{{20}} + k\pi \)\(\left( {k \in \mathbb{Z}} \right)\)