Bài tập 1.44 trang 40 SBT Toán 11

Ta có: cos3x−cos5x = sinx$$

\(\begin{array}{l}
 \Leftrightarrow \sin x + \cos 5x - \cos 3x = 0\\
 \Leftrightarrow \sin x - 2\sin \frac{{5x + 3x}}{2}\sin \frac{{5x - 3x}}{2} = 0\\
 \Leftrightarrow \sin x - 2\sin 4x\sin x = 0\\
 \Leftrightarrow \sin x(1 - 2\sin 4x) = 0\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\sin x = 0}\\
{\sin 4x = \frac{1}{2}}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = k\pi ,k \in Z}\\
{4x = \frac{\pi }{6} + k2\pi ,k \in Z}\\
{4x = \pi  - \frac{\pi }{6} + k2\pi ,k \in Z}
\end{array}} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi ,k \in Z\\
x = \frac{\pi }{{24}} + k\frac{\pi }{2},k \in Z\\
x = \frac{{5\pi }}{{24}} + k\frac{\pi }{2},k \in Z
\end{array} \right.
\end{array}\)

-- Mod Toán 11 HỌC247