Giải bài 59 trang 30 sách bài tập toán 11 – Cánh diều

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