Đáp án:
\(\left[ \matrix{
x = {10^0} + k{180^0} \hfill \cr
x = - {50^0} + k{180^0} \hfill \cr} \right.\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
$$\eqalign{
& \cos \left( {2x - {5^0}} \right) - \sin \left( {2x - {5^0}} \right) = {{\sqrt 2 } \over 2} \cr
& \Leftrightarrow {1 \over {\sqrt 2 }}\cos \left( {2x - {5^0}} \right) - \sin \left( {2x - {5^0}} \right) = {1 \over 2} \cr
& \Leftrightarrow \cos \left( {2x - {5^0}} \right)\cos {45^0} - \sin \left( {2x - {5^0}} \right)\sin {45^0} = {1 \over 2} \cr
& \Leftrightarrow \cos \left( {2x - {5^0} + {{45}^0}} \right) = {1 \over 2} \cr
& \Leftrightarrow \cos \left( {2x + {{40}^0}} \right) = {1 \over 2} \cr
& \Leftrightarrow \left[ \matrix{
2x + {40^0} = {60^0} + k{360^0} \hfill \cr
2x + {40^0} = - {60^0} + k{360^0} \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
2x = {20^0} + k{360^0} \hfill \cr
2x = - {100^0} + k{360^0} \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = {10^0} + k{180^0} \hfill \cr
x = - {50^0} + k{180^0} \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $$
