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) \(\cot x = \cot \left( { - {{32}^o}} \right)\)

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: \(\cot x = \cot \left( { - {{32}^o}} \right) \Leftrightarrow x =  - {32^o} + k{180^o}\)\(\left( {k \in \mathbb{Z}} \right)\)