Bài 5.49 trang 208 SBT đại số và giải tích 11

LG a

\(f'\left( x \right) = 0\) với \(f\left( x \right) = 1 - \sin \left( {\pi  + x} \right) + 2\cos {{3\pi  + x} \over 2}\)

Phương pháp giải:

Sử dụng công thức:

\(\begin{array}{l}
\sin \left( {\pi + \alpha } \right) = - \sin \alpha \\
\cos \left( { - \alpha } \right) = \cos \alpha \\
\cos \left( {\frac{\pi }{2} - \alpha } \right) = \sin \alpha
\end{array}\)

Lời giải chi tiết:

\(\begin{array}{l}
f\left( x \right) = 1 - \sin \left( {\pi + x} \right) + 2\cos \left( {\frac{{3\pi + x}}{2}} \right)\\
= 1 - \left( { - \sin x} \right) + 2\cos \left( {\frac{{3\pi }}{2} + \frac{x}{2}} \right)\\
 = 1 + \sin x + 2\cos \left( {2\pi  - \frac{\pi }{2} + \frac{x}{2}} \right)\\ = 1 + \sin x + 2\cos \left[ { - \left( {\frac{\pi }{2} - \frac{x}{2}} \right)} \right] \\= 1 + \sin x + 2\cos \left( {\frac{\pi }{2} - \frac{x}{2}} \right)\\= 1 + \sin x + 2\sin \frac{x}{2}\\
\Rightarrow f'\left( x \right) = \cos x + \cos \frac{x}{2}\\
f'\left( x \right) = 0 \Leftrightarrow \cos x + \cos \frac{x}{2} = 0\\
\Leftrightarrow 2{\cos ^2}\frac{x}{2} - 1 + \cos \frac{x}{2} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos \frac{x}{2} = - 1\\
\cos \frac{x}{2} = \frac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\frac{x}{2} = \pi + k2\pi \\
\frac{x}{2} = \frac{\pi }{3} + k2\pi \\
\frac{x}{2} = - \frac{\pi }{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\pi + k4\pi \\
x = \frac{{2\pi }}{3} + k4\pi \\
x = - \frac{{2\pi }}{3} + k4\pi
\end{array} \right.
\end{array}\)

LG b

\(g'\left( x \right) = 0\) với \(g\left( x \right) = \sin 3x - \sqrt 3 \cos 3x + 3\left( {\cos x - \sqrt 3 \sin x} \right).\)

Lời giải chi tiết:

\(\begin{array}{l}
g'\left( x \right) = 3\cos 3x + 3\sqrt 3 \sin 3x + 3\left( { - \sin x - \sqrt 3 \cos x} \right)\\
= 3\left( {\cos 3x + \sqrt 3 \sin 3x} \right) - 3\left( {\sin x + \sqrt 3 \cos x} \right)\\
g'\left( x \right) = 0\\
\Leftrightarrow 3\left( {\cos 3x + \sqrt 3 \sin 3x} \right) - 3\left( {\sin x + \sqrt 3 \cos x} \right) = 0\\
\Leftrightarrow \cos 3x + \sqrt 3 \sin 3x = \sin x + \sqrt 3 \cos x\\
\Leftrightarrow \frac{1}{2}\cos 3x + \frac{{\sqrt 3 }}{2}\sin 3x = \frac{1}{2}\sin x + \frac{{\sqrt 3 }}{2}\cos x\\
\Leftrightarrow \cos \left( {3x - \frac{\pi }{3}} \right) = \cos \left( {x - \frac{\pi }{6}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
3x - \frac{\pi }{3} = x - \frac{\pi }{6} + k2\pi \\
3x - \frac{\pi }{3} = - x + \frac{\pi }{6} + k2\pi 
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \frac{\pi }{6} + k2\pi \\
4x = \frac{\pi }{2} + k2\pi 
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{{12}} + k\pi \\
x = \frac{\pi }{8} + \frac{{k\pi }}{2}
\end{array} \right.
\end{array}\)

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