Câu hỏi 5 - Mục Bài tập trang 31

1. Nội dung câu hỏi

Tìm các nghiệm của mỗi phương trình sau trong khoảng \(\left( { - \pi ;\pi } \right)\).

a) \(\sin \left( {3x - \frac{\pi }{3}} \right) = 1\);

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

c) \(\tan \left( {x + \frac{\pi }{9}} \right) = \tan \frac{{4\pi }}{9}\).

3. Lời giải chi tiết 

a) \(\sin \left( {3x - \frac{\pi }{3}} \right) = 1 \) \( \Leftrightarrow 3x - \frac{\pi }{3} = \frac{\pi }{2} + k2\pi \left( {k \in \mathbb{Z}} \right) \) \( \Leftrightarrow x = \frac{{5\pi }}{{18}} + \frac{{k2\pi }}{3}\left( {k \in \mathbb{Z}} \right)\)

Vì \(x \in \left( { - \pi ;\pi } \right) \Rightarrow  - \pi  < \frac{{5\pi }}{{18}} + \frac{{k2\pi }}{3} < \pi  \) \( \Leftrightarrow \frac{{ - 23}}{{12}} < k < \frac{{13}}{{12}}\)

Mà \(k \in \mathbb{Z}\) nên \(k \in \left\{ { - 1;0;1} \right\}\). Do đó, \(x \in \left\{ {\frac{{ - 7\pi }}{{18}};\frac{{5\pi }}{{18}};\frac{{17\pi }}{{18}}} \right\}\).

b) \(2\cos \left( {2x - \frac{{3\pi }}{4}} \right) = \sqrt 3  \) \( \Leftrightarrow \cos \left( {2x - \frac{{3\pi }}{4}} \right) = \frac{{\sqrt 3 }}{2} \) \( \Leftrightarrow \cos \left( {2x - \frac{{3\pi }}{4}} \right) = \cos \frac{\pi }{6}\)

\( \Leftrightarrow \left[ \begin{array}{l}2x - \frac{{3\pi }}{4} = \frac{\pi }{6} + k2\pi \\2x - \frac{{3\pi }}{4} =  - \frac{\pi }{6} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right) \) \( \Leftrightarrow \left[ \begin{array}{l}x = \frac{{11\pi }}{{24}} + k\pi \\x = \frac{{7\pi }}{{24}} + k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

Vì \(x \in \left( { - \pi ;\pi } \right)\) nên:

TH1: \( - \pi  < \frac{{11\pi }}{{24}} + k\pi  < \pi  \) \( \Leftrightarrow \frac{{ - 35}}{{24}} < k < \frac{{13}}{{24}}\).

Mà \(k \in \mathbb{Z}\) nên \(k \in \left\{ { - 1;0} \right\}\). Do đó, \(x \in \left\{ {\frac{{ - 13\pi }}{{24}};\frac{{11\pi }}{{24}}} \right\}\).

TH2: \( - \pi  < \frac{{7\pi }}{{24}} + k\pi  < \pi  \) \( \Leftrightarrow \frac{{ - 31}}{{24}} < k < \frac{{17}}{{24}}\).

Mà \(k \in \mathbb{Z}\) nên \(k \in \left\{ { - 1;0} \right\}\). Do đó, \(x \in \left\{ {\frac{{ - 17\pi }}{{24}};\frac{{7\pi }}{{24}}} \right\}\).

Vậy \(x \in \left\{ {\frac{{ - 17\pi }}{{24}};\frac{{ - 13\pi }}{{24}};\frac{{7\pi }}{{24}};\frac{{11\pi }}{{24}}} \right\}\).

c) \(\tan \left( {x + \frac{\pi }{9}} \right) = \tan \frac{{4\pi }}{9} \) \( \Leftrightarrow x + \frac{\pi }{9} = \frac{{4\pi }}{9} + k\pi \left( {k \in \mathbb{Z}} \right) \) \( \Leftrightarrow x = \frac{\pi }{3} + k\pi \left( {k \in \mathbb{Z}} \right)\).

Vì \(x \in \left( { - \pi ;\pi } \right) \Rightarrow  - \pi  < \frac{\pi }{3} + k\pi  < \pi  \) \( \Leftrightarrow \frac{{ - 4}}{3} < k < \frac{2}{3}\)

Mà \(k \in \mathbb{Z}\) nên \(k \in \left\{ { - 1;0} \right\}\). Do đó, \(x \in \left\{ {\frac{{ - 2\pi }}{3};\frac{\pi }{3}} \right\}\).