Cho hàm số $y=f\left( x \right)$, hàm số $y={f}'\left( x \right)$...

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