Đề bài
Tìm góc lượng giác \(x\) sao cho:
a) \(\sin 2x = \sin {42^o}\)
b) \(\sin \left( {x - {{60}^o}} \right) = - \frac{{\sqrt 3 }}{2}\)
c) \(\cos \left( {x + {{50}^o}} \right) = \frac{1}{2}\)
d) \(\cos 2x = \cos \left( {3x + {{10}^o}} \right)\)
e) \(\tan x = \tan {25^o}\)
g) \(\cot x = \cot \left( { - {{32}^o}} \right)\)
Phương pháp giải - Xem chi tiết
Sử dụng các kết quả sau:
- \(\sin x = \sin \alpha \Leftrightarrow \left[ \begin{array}{l}x = \alpha + k{360^o}\\x = {180^o} - \alpha + k{360^o}\end{array} \right.\)\(\left( {k \in \mathbb{Z}} \right)\)
- \(\cos x = \cos \alpha \Leftrightarrow \left[ \begin{array}{l}x = \alpha + k{360^o}\\x = - \alpha + k{360^o}\end{array} \right.\)\(\left( {k \in \mathbb{Z}} \right)\)
- \(\tan x = \tan \alpha \Leftrightarrow x = \alpha + k{180^o}\)\(\left( {k \in \mathbb{Z}} \right)\)
- \(\cot x = \cot \alpha \Leftrightarrow x = \alpha + k{180^o}\)\(\left( {k \in \mathbb{Z}} \right)\)
Lời giải chi tiết
a) Ta có: \(\sin 2x = \sin {42^o} \Leftrightarrow \left[ \begin{array}{l}2x = {42^o} + k{360^o}\\2x = {180^o} - {42^o} + k{360^o}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = {21^o} + k{180^o}\\x = {69^o} + k{180^o}\end{array} \right.\)\(\left( {k \in \mathbb{Z}} \right)\)
b) Ta có \(\sin \left( { - {{60}^o}} \right) = - \frac{{\sqrt 3 }}{2}\), phương trình trở thành:
\(\sin \left( {x - {{60}^o}} \right) = \sin \left( { - {{60}^o}} \right) \Leftrightarrow \left[ \begin{array}{l}x - {60^o} = - {60^o} + k{360^o}\\x - {60^o} = {180^o} + {60^o} + k{360^o}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k{360^o}\\x = - {60^o} + k{360^o}\end{array} \right.\)\(\left( {k \in \mathbb{Z}} \right)\)
c) Ta có \(\cos {60^o} = \frac{1}{2}\), phương trình trở thành:
\(\cos \left( {x + {{50}^o}} \right) = \cos \left( {{{60}^o}} \right) \Leftrightarrow \left[ \begin{array}{l}x + {50^o} = {60^o} + k{360^o}\\x + {50^o} = - {60^o} + k{360^o}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = {10^o} + k{360^o}\\x = - {110^o} + k{360^o}\end{array} \right.\)\(\left( {k \in \mathbb{Z}} \right)\)
d) Ta có:
\(\cos 2x = \cos \left( {3x + {{10}^o}} \right) \Leftrightarrow \left[ \begin{array}{l}2x = 3x + {10^o} + k{360^o}\\2x = - \left( {3x + {{10}^o}} \right) + k{360^o}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - x = {10^o} + k{360^o}\\5x = - {10^o} + k{360^o}\end{array} \right.\)
\( \Leftrightarrow \left[ \begin{array}{l}x = - {10^o} + k{360^o}\\x = - {2^o} + k{72^o}\end{array} \right.\)\(\left( {k \in \mathbb{Z}} \right)\)
e) Ta có: \(\tan x = \tan {25^o} \Leftrightarrow x = {25^o} + k{180^o}\)\(\left( {k \in \mathbb{Z}} \right)\)
g) Ta có: \(\cot x = \cot \left( { - {{32}^o}} \right) \Leftrightarrow x = - {32^o} + k{180^o}\)\(\left( {k \in \mathbb{Z}} \right)\)
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