Giải bài 7 trang 169 – SGK môn Đại số và Giải tích lớp 11

a)

Ta có:

$\begin{aligned} & f'\left( x \right)=-3\sin x+4\cos x+5 \\ & \Rightarrow f'\left( x \right)=0 \\ & \Leftrightarrow -3\sin x+4\cos x+5=0 \\ & \Leftrightarrow 3\sin x-4\cos x=5 \\ & \Leftrightarrow \dfrac{4}{5}\cos x-\dfrac{3}{5}\sin x=-1 \\ & \Leftrightarrow \cos \left( \alpha +x \right)=-1 \\ & \Leftrightarrow \alpha +x=\pi +k2\pi \\ & \Leftrightarrow x=\pi -\alpha +k2\pi ,\,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned}$

Trong đó $\cos \alpha =\dfrac{4}{5};\,\sin \alpha =\dfrac{3}{5}$
b)

Ta có: 

$\begin{aligned} & f'\left( x \right)=-\cos \left( \pi +x \right)-\sin \left( \pi +\dfrac{x}{2} \right) \\ & \Rightarrow f'\left( x \right)=0 \\ & \Leftrightarrow -\cos \left( \pi +x \right)-\sin \left( \pi +\dfrac{x}{2} \right)=0 \\ & \Leftrightarrow \cos x+\sin \dfrac{x}{2}=0 \\ & \Leftrightarrow 1-2{{\sin }^{2}}\dfrac{x}{2}+\sin \dfrac{x}{2}=0 \\ & \Leftrightarrow \left[ \begin{aligned} & \sin \dfrac{x}{2}=1 \\ & \sin \dfrac{x}{2}=\dfrac{-1}{2} \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & \dfrac{x}{2}=\dfrac{\pi }{2}+k2\pi \\ & \dfrac{x}{2}=\dfrac{-\pi }{6}+k2\pi \\ & \dfrac{x}{2}=\pi +\dfrac{\pi }{6}+k2\pi \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x=\pi +k4\pi \\ & x=\dfrac{-\pi }{3}+k4\pi \\ & x=\dfrac{7\pi }{3}+k4\pi \\ \end{aligned} \right.\,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned}$