Phương pháp giải:

Biến đổi biểu thức, sử dụng công thức: $\sin \left( {\pi  + x} \right) =  - \sin x$, $\cos \left( {\dfrac{{3\pi }}{2} + \dfrac{x}{2}} \right) = \sin \dfrac{x}{2}$.

    Sử dụng công thức $\left( {\sin u} \right)' = u'\cos u$, $\left( {\cos u} \right)' =  - u'\sin u$.

Lời giải chi tiết:

$f\left( x \right) = 1 - \sin \left( {\pi  + x} \right) + 2\cos \left( {\dfrac{{3\pi }}{2} + \dfrac{x}{2}} \right)$.

$\begin{array}{l}f\left( x \right) = 1 - \left( { - \sin x} \right) + 2\sin \left( {\dfrac{x}{2}} \right)\\f\left( x \right) = 1 + \sin x + 2\sin \left( {\dfrac{x}{2}} \right)\\ \Rightarrow f'\left( x \right) = \cos x + \cos \left( {\dfrac{x}{2}} \right)\\f'\left( x \right) = 0 \Leftrightarrow \cos x + \cos \left( {\dfrac{x}{2}} \right) = 0\\ \Leftrightarrow \cos x =  - \cos \left( {\dfrac{x}{2}} \right)\\ \Leftrightarrow \cos x = \cos \left( {\pi  + \dfrac{x}{2}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{2} + x + k2\pi \\x =  - \dfrac{\pi }{2} - x + k2\pi \end{array} \right.\\ \Leftrightarrow x =  - \dfrac{\pi }{4} + k\pi \,\,\left( {k \in \mathbb{Z}} \right)\end{array}$

Vậy $x =  - \dfrac{\pi }{4} + k\pi \,\,\left( {k \in \mathbb{Z}} \right)$.

$\begin{array}{l}y = \dfrac{1}{2}\left( {\cos 8x + \cos 2x} \right)\\ \Rightarrow y' = \dfrac{1}{2}\left( { - 8\sin 8x - 2\sin 2x} \right)\\\,\,\,\,\,\,y' =  - 4\sin 8x - \sin 2x\end{array}$

Chọn C.

Phương pháp giải:

Sử dụng công thức $\left( {\frac{u}{v}} \right)' = \frac{{u'v - uv'}}{{{v^2}}}$.

Lời giải chi tiết:

$\begin{array}{l}f'\left( x \right) = 3\cos 3x + 3\sqrt 3 \sin 3x - 3\sin x - 3\sqrt 3 \cos x\\f'\left( x \right) = 0\\ \Leftrightarrow 3\cos 3x + 3\sqrt 3 \sin 3x - 3\sin x - 3\sqrt 3 \cos x = 0\\ \Leftrightarrow \sqrt 3 \sin 3x + \cos 3x = \sin x + \sqrt 3 \cos x\\ \Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin 3x + \dfrac{1}{2}\cos 3x = \dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x\\ \Leftrightarrow \sin 3x\cos \dfrac{\pi }{6} + \cos 3x\sin \dfrac{\pi }{6} = \sin x\cos \dfrac{\pi }{3} + \cos x\sin \dfrac{\pi }{3}\\ \Leftrightarrow \sin \left( {3x + \dfrac{\pi }{6}} \right) = \sin \left( {x + \dfrac{\pi }{3}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}3x + \dfrac{\pi }{6} = x + \dfrac{\pi }{3} + k2\pi \\3x + \dfrac{\pi }{6} = \pi  - x - \dfrac{\pi }{3} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}2x = \dfrac{\pi }{6} + k2\pi \\4x = \dfrac{\pi }{2} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{{12}} + k\pi \\x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}$

Vậy $\left[ \begin{array}{l}x = \dfrac{\pi }{{12}} + k\pi \\x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)$.

Phương pháp giải:

Sử dụng công thức $\left( {\frac{u}{v}} \right)' = \frac{{u'v - uv'}}{{{v^2}}}$.

Lời giải chi tiết:

$f\left( x \right) = 1 - {\sin ^4}3x + \dfrac{1}{6}\cos 6x$

$\begin{array}{l}f'\left( x \right) =  - 4{\sin ^3}3x.\left( {\sin 3x} \right)' - \sin 6x\\f'\left( x \right) =  - 12{\sin ^3}3x.\cos 3x - 2\sin 3x\cos 3x\\f'\left( x \right) =  - 2\sin 3x\cos 3x\left( {6{{\sin }^2}3x + 1} \right) = 0\\f'\left( x \right) =  - \sin 6x.\left[ {3\left( {1 - \cos 6x} \right) + 1} \right] = 0\\f'\left( x \right) =  - \sin 6x\left( {4 - 3\cos 6x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin 6x = 0\\\cos 6x = \dfrac{4}{3}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}6x = k\pi \\6x =  \pm \arccos \dfrac{4}{3} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{{k\pi }}{6}\\x =  \pm \dfrac{1}{6}\arccos \dfrac{4}{3} + \dfrac{{k\pi }}{3}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}$

Vậy $x = \dfrac{{k\pi }}{6};\,\,x =  \pm \dfrac{1}{6}\arccos \dfrac{4}{3} + \dfrac{{k\pi }}{3}\,\,\left( {k \in \mathbb{Z}} \right)$.

Chọn B.