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Câu hỏi: Giải các phương trình sau:
a) \(\sin x = \frac{{\sqrt 3 }}{2}\);
b) \(2\cos x = - \sqrt 2 \);
c) \(\sqrt 3 \tan \left( {\frac{x}{2} + {{15}^0}} \right) = 1\);
d) \(\cot \left( {2x + 1} \right) = \cot \frac{\pi }{5}\)
a) \(\sin x = \frac{{\sqrt 3 }}{2}\);
b) \(2\cos x = - \sqrt 2 \);
c) \(\sqrt 3 \tan \left( {\frac{x}{2} + {{15}^0}} \right) = 1\);
d) \(\cot \left( {2x + 1} \right) = \cot \frac{\pi }{5}\)
Phương pháp giải
Dựa vào công thức nghiệm tổng quát:
\(\sin x = m \Leftrightarrow \sin x = \sin \alpha \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \alpha + k2\pi }\\{x = \pi - \alpha + k2\pi }\end{array}\left( {k \in \mathbb{Z}} \right)} \right.\)
\(\cos x = m \Leftrightarrow \cos x = \cos \alpha \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \alpha + k2\pi }\\{x = - \alpha + k2\pi }\end{array} \left( {k \in \mathbb{Z}} \right)} \right. \)
\(\tan x = m \Leftrightarrow \tan x = \tan \alpha \Leftrightarrow x = \alpha + k\pi \left( {k \in \mathbb{Z}} \right)\)
\(\cot x = m \Leftrightarrow \cot x = \cot \alpha \Leftrightarrow x = \alpha + k\pi \left( {k \in \mathbb{Z}} \right)\)
Lời giải chi tiết
a) \(\sin x = \frac{{\sqrt 3 }}{2} \Leftrightarrow \sin x = \sin \frac{\pi }{3} \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \pi - \frac{\pi }{3} + k2\pi }\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \frac{{2\pi }}{3} + k2\pi }\end{array} } \right.\left( {k \in \mathbb{Z}} \right)\)
b) \(2\cos x = - \sqrt 2 \Leftrightarrow \cos x = - \frac{{\sqrt 2 }}{2} \Leftrightarrow \cos x = \cos \frac{{3\pi }}{4} \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{{3\pi }}{4} + k2\pi }\\{x = - \frac{{3\pi }}{4} + k2\pi }\end{array} \left( {k \in \mathbb{Z}} \right)} \right.\)
c) \(\sqrt 3 \left( {\tan \frac{x}{2} + {{15}^0}} \right) = 1 \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \frac{1}{{\sqrt 3 }} \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \tan \frac{\pi }{6}\)
\( \Leftrightarrow \frac{x}{2} + \frac{\pi }{{12}} = \frac{\pi }{6} + k\pi \Leftrightarrow \frac{x}{2} = \frac{\pi }{{12}} + k\pi \Leftrightarrow x = \frac{\pi }{6} + k\pi \left( {k \in \mathbb{Z}} \right)\)
d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5} \Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2} \left( {k \in \mathbb{Z}} \right)\)
Dựa vào công thức nghiệm tổng quát:
\(\sin x = m \Leftrightarrow \sin x = \sin \alpha \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \alpha + k2\pi }\\{x = \pi - \alpha + k2\pi }\end{array}\left( {k \in \mathbb{Z}} \right)} \right.\)
\(\cos x = m \Leftrightarrow \cos x = \cos \alpha \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \alpha + k2\pi }\\{x = - \alpha + k2\pi }\end{array} \left( {k \in \mathbb{Z}} \right)} \right. \)
\(\tan x = m \Leftrightarrow \tan x = \tan \alpha \Leftrightarrow x = \alpha + k\pi \left( {k \in \mathbb{Z}} \right)\)
\(\cot x = m \Leftrightarrow \cot x = \cot \alpha \Leftrightarrow x = \alpha + k\pi \left( {k \in \mathbb{Z}} \right)\)
Lời giải chi tiết
a) \(\sin x = \frac{{\sqrt 3 }}{2} \Leftrightarrow \sin x = \sin \frac{\pi }{3} \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \pi - \frac{\pi }{3} + k2\pi }\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \frac{{2\pi }}{3} + k2\pi }\end{array} } \right.\left( {k \in \mathbb{Z}} \right)\)
b) \(2\cos x = - \sqrt 2 \Leftrightarrow \cos x = - \frac{{\sqrt 2 }}{2} \Leftrightarrow \cos x = \cos \frac{{3\pi }}{4} \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{{3\pi }}{4} + k2\pi }\\{x = - \frac{{3\pi }}{4} + k2\pi }\end{array} \left( {k \in \mathbb{Z}} \right)} \right.\)
c) \(\sqrt 3 \left( {\tan \frac{x}{2} + {{15}^0}} \right) = 1 \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \frac{1}{{\sqrt 3 }} \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \tan \frac{\pi }{6}\)
\( \Leftrightarrow \frac{x}{2} + \frac{\pi }{{12}} = \frac{\pi }{6} + k\pi \Leftrightarrow \frac{x}{2} = \frac{\pi }{{12}} + k\pi \Leftrightarrow x = \frac{\pi }{6} + k\pi \left( {k \in \mathbb{Z}} \right)\)
d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5} \Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2} \left( {k \in \mathbb{Z}} \right)\)