Giải phương trình \(f'\left( x \right) = g\left( x \right),\) biết rằng:
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