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)$