Giải bài 5.113 trang 217 - SBT Đại số và Giải tích lớp 11

a)

Ta có: 

$f'\left( x \right)=\sin 3x$

Suy ra:

$\begin{aligned} & f'\left( x \right)=g\left( x \right) \\ & \Leftrightarrow \sin 3x=\left( \cos 6x-1 \right)\cot 3x \\ & \Leftrightarrow \sin 3x=\left( \cos 6x-1 \right).\dfrac{\cos 3x}{\sin 3x} \\ & \Rightarrow {{\sin }^{2}}3x=-2\sin^23x.\cos 3x \\ & \Leftrightarrow \cos 3x=\dfrac {-1}{2}\\ &\Leftrightarrow 3x=\pm \dfrac{2\pi}{3}+k2\pi \\ &\Leftrightarrow x=\pm\dfrac{2\pi}{9}+k\dfrac{2\pi}{3}\,\,\,(k\in \mathbb Z) \\ \end{aligned}$

b)

$\begin{aligned} & f'\left( x \right)=-\sin 2x \\ & \Rightarrow f'\left( x \right)=g\left( x \right) \\ & \Leftrightarrow -\sin 2x=1-{{\left( \cos 3x+\sin 3x \right)}^{2}} \\ & \Leftrightarrow 1+\sin 2x=1+\sin 6x \\ & \Leftrightarrow \sin 2x=\sin 6x \\ & \Leftrightarrow \left[ \begin{aligned} & 6x=2x+k2\pi \\ & 6x=\pi -2x+k2\pi \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & 4x=k2\pi \\ & 8x=\pi +k2\pi \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x=\frac{k\pi }{2} \\ & x=\frac{\pi }{8}+k\frac{\pi }{4} \\ \end{aligned} \right.\,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned}$

c)

$\begin{aligned} & f'\left( x \right)=\cos 2x-5\sin x \\ & \Rightarrow \cos 2x-5\sin x=3{{\sin }^{2}}x+\frac{3}{1+{{\tan }^{2}}x} \\ & \Leftrightarrow \cos 2x-5\sin x=3{{\sin }^{2}}x+3{{\cos }^{2}}x \\ & \Leftrightarrow \cos 2x-5\sin x-3=0 \\ & \Leftrightarrow -2{{\sin }^{2}}x-5\sin x-2=0 \\ & \Leftrightarrow \left[ \begin{aligned} & \sin x=-\frac{1}{2} \\ & \sin x=-2\,\,\,\,\left( \text{loại} \right) \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x=-\frac{\pi }{6}+k2\pi \\ & x=\frac{7\pi }{6}+k2\pi \\ \end{aligned} \right.\,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned}$