Bài tập 26 trang 32 SGK Toán 11 NC

a)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\cos 3x = \sin 2x\\
 \Leftrightarrow \cos 3x - \cos \left( {\frac{\pi }{2} - 2x} \right) = 0
\end{array}\\
{ \Leftrightarrow  - 2\sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right)\sin \left( {\frac{{5x}}{2} - \frac{\pi }{4}} \right) = 0}\\
{ \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\frac{x}{2} + \frac{\pi }{4} = k\pi }\\
{\frac{{5x}}{2} - \frac{\pi }{4} = k\pi }
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x =  - \frac{\pi }{2} + k2\pi }\\
{x = \frac{\pi }{{10}} + k\frac{{2\pi }}{5}}
\end{array}} \right.}
\end{array}\)

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\sin \left( {x - {{120}^0}} \right) - \cos 2x = 0\\
 \Leftrightarrow \cos \left( {{{210}^0} - x} \right) - \cos 2x = 0
\end{array}\\
{ \Leftrightarrow  - 2\sin \left( {\frac{x}{2} + {{105}^0}} \right)\sin \left( {{{105}^0} - \frac{{3x}}{2}} \right) = 0}\\
{ \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\frac{x}{2} + {{105}^0} = k{{180}^0}}\\
{{{105}^0} - \frac{{3x}}{2} = k{{180}^0}}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x =  - {{210}^0} + k{{360}^0}}\\
{x = {{70}^0} + k{{120}^0}}
\end{array}} \right.}
\end{array}\)

-- Mod Toán 11 HỌC247