Bài tập 58 trang 30 SBT Toán 11 Tập 1 Cánh diều - CD

a) Ta có: \(\sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2}\) nên 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: \(\sin \left( { - \frac{\pi }{4}} \right) = - \frac{{\sqrt 2 }}{2}\) nên 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: \(\cos \frac{{2\pi }}{3} = \frac{{ - 1}}{2}\) nên phương trình trở thành:

\(\cos \left( {3x + \frac{\pi }{3}} \right) = \cos \frac{{2\pi }}{3} \\\Leftrightarrow \left[ \begin{array}{l}3x + \frac{\pi }{3} = \frac{{2\pi }}{3} + k2\pi \\3x + \frac{\pi }{3} = - \frac{{2\pi }}{3} + k2\pi \end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l}3x = \frac{\pi }{3} + k2\pi \\3x = - \pi + k2\pi \end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{9} + k\frac{{2\pi }}{3}\\x = - \frac{\pi }{3} + k\frac{{2\pi }}{3}\end{array} \right.\)

-- Mod Toán 11 HỌC247