Cos(2x-5°)-sin(2x-5°)=((Sqrt(2))/(2))

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:

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