- LG a
- LG b
Giải các phương trình
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:
\[\begin{array}{l}
g'\left[ x \right] = 3\cos 3x + 3\sqrt 3 \sin 3x + 3\left[ { - \sin x - \sqrt 3 \cos x} \right]\\
= 3\left[ {\cos 3x + \sqrt 3 \sin 3x} \right] - 3\left[ {\sin x + \sqrt 3 \cos x} \right]\\
g'\left[ x \right] = 0\\
\Leftrightarrow 3\left[ {\cos 3x + \sqrt 3 \sin 3x} \right] - 3\left[ {\sin x + \sqrt 3 \cos x} \right] = 0\\
\Leftrightarrow \cos 3x + \sqrt 3 \sin 3x = \sin x + \sqrt 3 \cos x\\
\Leftrightarrow \frac{1}{2}\cos 3x + \frac{{\sqrt 3 }}{2}\sin 3x = \frac{1}{2}\sin x + \frac{{\sqrt 3 }}{2}\cos x\\
\Leftrightarrow \cos \left[ {3x - \frac{\pi }{3}} \right] = \cos \left[ {x - \frac{\pi }{6}} \right]\\
\Leftrightarrow \left[ \begin{array}{l}
3x - \frac{\pi }{3} = x - \frac{\pi }{6} + k2\pi \\
3x - \frac{\pi }{3} = - x + \frac{\pi }{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \frac{\pi }{6} + k2\pi \\
4x = \frac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{{12}} + k\pi \\
x = \frac{\pi }{8} + \frac{{k\pi }}{2}
\end{array} \right.
\end{array}\]