Bài tập 59 trang 30 SBT Toán 11 Tập 1 Cánh diều - CD

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: \(\sin \left( { - {{60}^o}} \right) = - \frac{{\sqrt 3 }}{2}\) nên 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: \(\tan x = \tan {25^o} \Leftrightarrow x = {25^o} + k{180^o}\)\(\left( {k \in \mathbb{Z}} \right)\)

-- Mod Toán 11 HỌC247