Bài tập 73 trang 33 SBT Toán 11 Tập 1 Cánh diều - CD

a) Ta có: \(\sin \left( {2x + \frac{\pi }{3}} \right) = \sin \left( {3x - \frac{\pi }{6}} \right)\)

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

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

b) Ta có: \(\cos \left( {x + \frac{\pi }{4}} \right) = \cos \left( {\frac{\pi }{4} - 2x} \right)\)

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

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

c) Ta có: \({\cos ^2}\left( {\frac{x}{2} + \frac{\pi }{6}} \right) = \frac{{1 + \cos \left[ {2\left( {\frac{x}{2} + \frac{\pi }{6}} \right)} \right]}}{2} = \frac{{1 + \cos \left( {x + \frac{\pi }{3}} \right)}}{2}\);

             \({\cos ^2}\left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{{1 + \cos \left[ {2\left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right)} \right]}}{2} = \frac{{1 + \cos \left( {3x + \frac{\pi }{2}} \right)}}{2}\).

Phương trình trở thành: \(\frac{{1 + \cos \left( {x + \frac{\pi }{3}} \right)}}{2} = \frac{{1 + \cos \left( {3x + \frac{\pi }{2}} \right)}}{2}\)

\(\Leftrightarrow \cos \left( {x + \frac{\pi }{3}} \right) = \cos \left( {3x + \frac{\pi }{2}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}x + \frac{\pi }{3} = 3x + \frac{\pi }{2} + k2\pi \\x + \frac{\pi }{3} = - 3x - \frac{\pi }{2} + k2\pi \end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l} - 2x = \frac{\pi }{6} + k2\pi \\4x = - \frac{{5\pi }}{6} + k2\pi \end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{{12}} + k\pi \\x = - \frac{{5\pi }}{{24}} + k\frac{\pi }{2}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

d) Ta có: \(\tan \frac{{2\pi }}{7} = \cot \left( {\frac{\pi }{2} - \frac{{2\pi }}{7}} \right) = \cot \frac{{3\pi }}{{14}}\).

Phương trình trở thành: \(\cot 3x = \cot \frac{{3\pi }}{{14}} \Leftrightarrow 3x = \frac{{3\pi }}{{14}} + k\pi \Leftrightarrow x = \frac{\pi }{{14}} + k\frac{\pi }{3}\)\(\left( {k \in \mathbb{Z}} \right)\)

-- Mod Toán 11 HỌC247