Bài 72 trang 32 SBT toán 11 – Cánh diều: Giải phương trình: a) sin 2x – π /6 = – 1/2 b) sin x/3 + π /2 = √3 /2

Đề bài/câu hỏi:

Giải phương trình:

a) \(\sin \left( {2x – \frac{\pi }{6}} \right) = – \frac{1}{2}\)

b) \(\sin \left( {\frac{x}{3} + \frac{\pi }{2}} \right) = \frac{{\sqrt 3 }}{2}\)

c) \(\cos \left( {2x + \frac{\pi }{5}} \right) = \frac{{\sqrt 2 }}{2}\)

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

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

g) \(\cot \left( {3x + \pi } \right) = – 1\)

Hướng dẫn:

Sử dụng các kết quả sau:

Lời giải:

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:

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

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:

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