Bài tập 5.113 trang 217 SBT Toán 11

a) Ta có: \(f'\left( x \right) = \sin 3x\)

Suy ra:

\(\begin{array}{l}
f'\left( x \right) = g\left( x \right) \Leftrightarrow \sin 3x = \left( {\cos 6x - 1} \right)\\
 \Leftrightarrow \sin 3x = \left( {\cos 6x - 1} \right).\frac{{\cos 3x}}{{\sin 3x}}\\
 \Rightarrow {\sin ^2}3x =  - 2{\sin ^2}3x.\cos 3x\\
 \Leftrightarrow \cos 3x =  - \frac{1}{2}\\
 \Leftrightarrow 3x =  \pm \frac{{2\pi }}{3} + k2\pi \\
 \Leftrightarrow x =  \pm \frac{{2\pi }}{9} + k\frac{{2\pi }}{3}\left( {k \in Z} \right)
\end{array}\)

b)

\(\begin{array}{l}
f\prime \left( x \right) =  - \sin 2x\\
 \Rightarrow f\prime \left( x \right) = g\left( x \right) \Leftrightarrow  - \sin 2x = 1 - {\left( {\cos 3x + \sin 3x} \right)^2}\\
 \Leftrightarrow 1 + \sin 2x = 1 + \sin 6x \Leftrightarrow \sin 2x = \sin 6x\\
 \Leftrightarrow \left[ \begin{array}{l}
6x = 2x + k2\pi \\
6x = \pi  - 2x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
4x = k2\pi \\
8x = \pi  + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{k\pi }}{2}\\
x = \frac{\pi }{8} + k\frac{\pi }{4}
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)

c)

\(\begin{array}{l}
f\prime (x) = \cos 2x - 5\sin x\\
 \Rightarrow \cos 2x - 5\sin x = 3{\sin ^2}x + \frac{3}{{1 + {{\tan }^2}x}}\\
 \Leftrightarrow \cos 2x - 5\sin x = 3{\sin ^2}x + 3{\cos ^2}x\\
 \Leftrightarrow \cos 2x - 5\sin x - 3 = 0\\
 \Leftrightarrow  - 2{\sin ^2}x - 5\sin x - 2 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x =  - \frac{1}{2}\\
\sin x =  - 2\,\,\left( l \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \frac{\pi }{6} + k2\pi \\
x = \frac{{7\pi }}{6} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)

-- Mod Toán 11 HỌC247