Bài 1.34 trang 41 SGK Toán 11 tập 1 - Kết nối tri thức

Đề bài

Giải các phương trình sau:

a) \(\cos \left( {3x - \frac{\pi }{4}} \right) =  - \frac{{\sqrt 2 }}{2}\);        

b) \(2{\sin ^2}x - 1 + \cos 3x = 0\);        

c) \(\tan \left( {2x + \frac{\pi }{5}} \right) = \tan \left( {x - \frac{\pi }{6}} \right)\).

Lời giải chi tiết

a) \(\cos \left( {3x - \frac{\pi }{4}} \right) =  - \frac{{\sqrt 2 }}{2} \)

\(\Leftrightarrow \cos \left( {3x - \frac{\pi }{4}} \right) = \cos \frac{{3\pi }}{4} \)

\(\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x - \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi }\\{3x - \frac{\pi }{4} =  - \frac{{3\pi }}{4} + k2\pi }\end{array}} \right. \)

\(\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x = \pi  + k2\pi }\\{3x =  - \frac{\pi }{2} + k2\pi }\end{array}} \right.\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + \frac{{k2\pi }}{3}}\\{x =  - \frac{\pi }{6} + \frac{{k2\pi }}{3}}\end{array}} \right.\) \(\left( {k \in \mathbb{Z}} \right)\).

b) \(2{\sin ^2}x - 1 + \cos 3x = 0\)

\( \Leftrightarrow -\cos 2x + \cos 3x = 0\)

\( \Leftrightarrow \cos 3x = \cos 2x\)

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

\( \Leftrightarrow \left[ \begin{array}{l}x = k2\pi \\5x = k2\pi \end{array} \right.\)

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

c) \(\tan \left( {2x + \frac{\pi }{5}} \right) = \tan \left( {x - \frac{\pi }{6}} \right)\)

\( \Leftrightarrow 2x + \frac{\pi }{5} = x - \frac{\pi }{6} + k\pi \)

\( \Leftrightarrow x =  - \frac{{11\pi }}{{30}} + k\pi \) \(\left( {k \in \mathbb{Z}} \right)\).