Lời giải:.jpg.png)
Ta có: \(g'\left( x \right) = 5\cos xf'\left( {\frac{{5\sin x - 1}}{2}} \right) + \frac{5}{2}\cos x\left( {5\sin x - 1} \right)\).
\(g'\left( x \right) = 0 \Leftrightarrow 5\cos xf'\left( {\frac{{5\sin x - 1}}{2}} \right) + \frac{5}{2}\cos x\left( {5\sin x - 1} \right) = 0\)
\( \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
f'\left( {\frac{{5\sin x - 1}}{2}} \right) = - \frac{{5\sin x - 1}}{2}
\end{array} \right.\)
\( \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\frac{{5\sin x - 1}}{2} = - 3\\
\frac{{5\sin x - 1}}{2} = - 1\\
\frac{{5\sin x - 1}}{2} = \frac{1}{3}\\
\frac{{5\sin x - 1}}{2} = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
5\sin x - 1 = - 6\\
5\sin x - 1 = - 2\\
5\sin x - 1 = \frac{2}{3}\\
5\sin x - 1 = 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin x = - 1\\
\sin x = - \frac{1}{5}\\
\sin x = \frac{1}{3}\\
\sin x = \frac{3}{5}
\end{array} \right.\)
\( \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin x = - 1\\
\sin x = - \frac{1}{5}\\
\sin x = \frac{1}{3}\\
\sin x = \frac{3}{5}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{2} \vee x = \frac{{3\pi }}{2}\\
x = \frac{{3\pi }}{2}\\
x = \pi - arc\sin \left( { - \frac{1}{5}} \right) \vee x = 2\pi + arc\sin \left( { - \frac{1}{5}} \right)\\
x = arc\sin \left( {\frac{1}{3}} \right) \vee x = \pi - arc\sin \left( {\frac{1}{3}} \right)\\
x = arc\sin \left( {\frac{3}{5}} \right) \vee x = \pi - arc\sin \left( {\frac{3}{5}} \right)
\end{array} \right.\)
Suy phương trình g'(x) = 0 có 9 nghiệm, trong đó có nghiệm \(x = \frac{{3\pi }}{2}\) là nghiệm kép.
Vậy hàm số y = g(x) có 7 cực trị.
