Bài 1 trang 40 SGK Toán 11 tập 1 – Cánh diều

Đề bài

Giải phương trình:

a)     \(\sin \left( {2x – \frac{\pi }{3}} \right) =  – \frac{{\sqrt 3 }}{2}\)

b)     \(\sin \left( {3x + \frac{\pi }{4}} \right) =  – \frac{1}{2}\)

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

d)     \(2\cos 3x + 5 = 3\)

e)     \(3\tan x =  – \sqrt 3 \)

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

Lời giải chi tiết

a)     \(\sin \left( {2x – \frac{\pi }{3}} \right) =  – \frac{{\sqrt 3 }}{2}\)

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

Vậy phương trình có nghiệm là: \(x \in \left\{ {k\pi ;\frac{{5\pi }}{6} + k\pi } \right\}\)

b)     \(\sin \left( {3x + \frac{\pi }{4}} \right) =  – \frac{1}{2}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}3x + \frac{\pi }{4} =  – \frac{\pi }{6} + k2\pi \\3x + \frac{\pi }{4} = \frac{{7\pi }}{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}3x =  – \frac{{5\pi }}{{12}} + k2\pi \\3x = \frac{{11\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x =  – \frac{{5\pi }}{{36}} + k\frac{{2\pi }}{3}\\x = \frac{{11\pi }}{{36}} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

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

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} + \frac{\pi }{4} = \frac{\pi }{6} + k2\pi \\\frac{x}{2} + \frac{\pi }{4} =  – \frac{\pi }{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} =  – \frac{\pi }{{12}} + k2\pi \\\frac{x}{2} =  – \frac{{5\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x =  – \frac{\pi }{6} + k4\pi \\x =  – \frac{{5\pi }}{6} + k4\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

d)     \(2\cos 3x + 5 = 3\)

\(\begin{array}{l} \Leftrightarrow \cos 3x =  – 1\\ \Leftrightarrow \left[ \begin{array}{l}3x = \pi  + k2\pi \\3x =  – \pi  + k2\pi \end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{3} + k\frac{{2\pi }}{3}\\x = \frac{{ – \pi }}{3} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

e)      

\(\begin{array}{l}3\tan x =  – \sqrt 3 \\ \Leftrightarrow \tan x = \frac{{ – \sqrt 3 }}{3}\\ \Leftrightarrow \tan x = \tan \left( { – \frac{\pi }{6}} \right)\\ \Leftrightarrow x =  – \frac{\pi }{6} + k\pi \end{array}\)

g)

\(\begin{array}{l}\cot x – 3 = \sqrt 3 \left( {1 – \cot x} \right)\\ \Leftrightarrow \cot x – 3 = \sqrt 3  – \sqrt 3 \cot x\\ \Leftrightarrow \cot x + \sqrt 3 \cot x = \sqrt 3  + 3\\ \Leftrightarrow (1 + \sqrt 3 )\cot x = \sqrt 3  + 3\\ \Leftrightarrow \cot x = \sqrt 3 \\ \Leftrightarrow \cot x = \cot \frac{\pi }{6}\\ \Leftrightarrow x = \frac{\pi }{6} + k\pi \end{array}\)