Giải bài 73 trang 33 sách bài tập toán 11 - Cánh diều — Không quảng cáo

Đề bài

Giải phương trình:

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

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

c) ${\cos ^2}\left( {\frac{x}{2} + \frac{\pi }{6}} \right) = {\cos ^2}\left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right)$

d) $\cot 3x = \tan \frac{{2\pi }}{7}$

Lời giải chi tiết

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)$