Giải phương trình \(f'\left( x \right) = g\left( x \right)\)
LG a
Với \(f\left( x \right) = 1 - {\sin ^4}3x\) và \(g\left( x \right) = \sin 6x\)
Lời giải chi tiết:
\(\begin{array}{l}
f'\left( x \right) = - 4{\sin ^3}3x.\left( {\sin 3x} \right)'\\
= - 4{\sin ^3}3x.3\cos 3x\\
= - 12{\sin ^3}3x\cos 3x\\= - 6{\sin ^2}3x.2\sin 3x\cos 3x\\
= - 6{\sin ^2}3x\sin 6x\\
f'\left( x \right) = g\left( x \right)\\
\Leftrightarrow - 6{\sin ^2}3x\sin 6x = \sin 6x\\
\Leftrightarrow \sin 6x\left( {1 + 6{{\sin }^2}3x} \right) = 0\\
\Leftrightarrow \sin 6x = 0\left( {do\,1 + 6{{\sin }^2}3x > 0} \right)\\
\Leftrightarrow 6x = k\pi \\
\Leftrightarrow x = \dfrac{{k\pi }}{6}
\end{array}\)
LG b
Với \(f\left( x \right) = 4x{\cos ^2}\left( {{x \over 2}} \right)\) và \(g\left( x \right) = 8\cos {x \over 2} - 3 - 2x\sin x.\)
Lời giải chi tiết:
\(\begin{array}{l}
f'\left( x \right) = 4{\cos ^2}\frac{x}{2} + 4x.2\cos \frac{x}{2}\left( {\cos \frac{x}{2}} \right)'\\
= 4{\cos ^2}\frac{x}{2} + 8x\cos \frac{x}{2}.\left( { - \frac{1}{2}\sin \frac{x}{2}} \right)\\
= 4{\cos ^2}\frac{x}{2} - 4x\cos \frac{x}{2}\sin \frac{x}{2}\\
= 4{\cos ^2}\frac{x}{2} - 2x\sin x\\
f'\left( x \right) = g\left( x \right)\\
\Leftrightarrow 4{\cos ^2}\frac{x}{2} - 2x\sin x \\= 8\cos \frac{x}{2} - 3 - 2x\sin x\\
\Leftrightarrow 4{\cos ^2}\frac{x}{2} - 8\cos \frac{x}{2} + 3 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos \frac{x}{2} = \frac{3}{2}\left( {VN} \right)\\
\cos \frac{x}{2} = \frac{1}{2}
\end{array} \right.\\
\Leftrightarrow \frac{x}{2} = \pm \frac{\pi }{3} + k2\pi \\
\Leftrightarrow x = \pm \frac{{2\pi }}{3} + k4\pi
\end{array}\)
