Bài tập 72 trang 32 SBT Toán 11 Tập 1 Cánh diều - CD

a) Ta có: \(\sin \left( { - \frac{\pi }{6}} \right) = - \frac{1}{2}\).

Phương trình trở thành: \(\sin \left( {2x - \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{6}} \right)\)

\(\Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{6} = - \frac{\pi }{6} + k2\pi \\2x - \frac{\pi }{6} = \pi + \frac{\pi }{6} + k2\pi \end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l}2x = k2\pi \\2x = \frac{{4\pi }}{3} + k2\pi \end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \frac{{2\pi }}{3} + k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

b) Ta có: \(\sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2}\).

Phương trình trở thành: \(\sin \left( {\frac{x}{3} + \frac{\pi }{2}} \right) = \sin \frac{\pi }{3}\)

\(\Leftrightarrow \left[ \begin{array}{l} + \frac{\pi }{2} = \frac{\pi }{3} + k2\pi \\\frac{x}{3} + \frac{\pi }{2} = \pi - \frac{\pi }{3} + k2\pi \end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l}\frac{x}{3} = - \frac{\pi }{6} + k2\pi \\\frac{x}{3} = \frac{\pi }{6} + k2\pi \end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{2} + k6\pi \\x = \frac{\pi }{2} + k6\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

c) Ta có: \(\cos \frac{\pi }{4} = \frac{{\sqrt 2 }}{2}\).

Phương trình trở thành: \(\cos \left( {2x + \frac{\pi }{5}} \right) = \cos \frac{\pi }{4}\)

\(\Leftrightarrow \left[ \begin{array}{l}2x + \frac{\pi }{5} = \frac{\pi }{4} + k2\pi \\2x + \frac{\pi }{5} = - \frac{\pi }{4} + k2\pi \end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{{20}} + k2\pi \\2x = - \frac{{9\pi }}{{20}} + k2\pi \end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{40}} + k\pi \\x = - \frac{{9\pi }}{{40}} + k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

d) \(2\cos \frac{x}{2} + \sqrt 3 = 0 \Leftrightarrow \cos \frac{x}{2} = - \frac{{\sqrt 3 }}{2}\)

Ta có: \(\cos \frac{{5\pi }}{6} = - \frac{{\sqrt 3 }}{2}\).

Phương trình trở thành: \(\cos \frac{x}{2} = \cos \frac{{5\pi }}{6}\)

\(\Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} = \frac{{5\pi }}{6} + k2\pi \\\frac{x}{2} = - \frac{{5\pi }}{6} + k2\pi \end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l}x = \frac{{5\pi }}{3} + k4\pi \\x = - \frac{{5\pi }}{3} + k4\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

e) \(\sqrt 3 \tan \left( {2x + \frac{\pi }{3}} \right) - 1 = 0 \Leftrightarrow \tan \left( {2x + \frac{\pi }{3}} \right) = \frac{1}{{\sqrt 3 }}\)

Ta có: \(\tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }}\).

Phương trình trở thành: \(\tan \left( {2x + \frac{\pi }{3}} \right) = \tan \frac{\pi }{6}\)

\(\Leftrightarrow 2x + \frac{\pi }{3} = \frac{\pi }{6} + k\pi \\\Leftrightarrow 2x = - \frac{\pi }{6} + k\pi \\\Leftrightarrow x = - \frac{\pi }{2} + k\frac{\pi }{2}\left( {k \in \mathbb{Z}} \right)\)

f) Ta có: \(\cot \left( { - \frac{\pi }{4}} \right) = - 1\).

Phương trình trở thành: \(\cot \left( {3x + \pi } \right) = \cot \frac{{ - \pi }}{4}\)

\(\Leftrightarrow 3x + \pi = \frac{{ - \pi }}{4} + k\pi \\\Leftrightarrow 3x = \frac{{ - \pi }}{4} + k\pi \\\Leftrightarrow x = \frac{{ - \pi }}{{12}} + k\frac{\pi }{3}\left( {k \in \mathbb{Z}} \right)\)

-- Mod Toán 11 HỌC247