Giải các phương trình :
LG a
\({\sin ^4}x + {\cos ^4}x = {3 \over 4}\)
Lời giải chi tiết:
LG b
\({\sin ^2}2x - {\sin ^2}x = {\sin ^2}{\pi \over 4}\)
Lời giải chi tiết:
\(\eqalign{ & {\sin ^2}2x - {\sin ^2}x = {\sin ^2}{\pi \over 4} \cr & \Leftrightarrow 4{\sin ^2}x{\cos ^2}x - {\sin ^2}x = {1 \over 2} \cr & \Leftrightarrow 8{\sin ^2}x\left( {1 - {{\sin }^2}x} \right) - 2{\sin ^2}x = 1 \cr & \Leftrightarrow 8{\sin ^4}x - 6{\sin ^2}x + 1 = 0 \cr & \Leftrightarrow \left[ {\matrix{ {{{\sin }^2}x = {1 \over 2}} \cr {{{\sin }^2}x = {1 \over 4}} \cr } } \right. \Leftrightarrow \left[ {\matrix{ {{{1 - \cos 2x} \over 2} = {1 \over 2}} \cr {{{1 - \cos 2x} \over 2} = {1 \over 4}} \cr } } \right. \cr & \Leftrightarrow \left[ {\matrix{ {\cos 2x = 0} \cr {\cos 2x = {1 \over 2}} \cr } } \right. \cr} \)
\( \Leftrightarrow \left[ \begin{array}{l}
2x = \frac{\pi }{2} + k\pi \\
2x = \pm \frac{\pi }{3} + k2\pi
\end{array} \right. \) \(\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{4} + \frac{{k\pi }}{2}\\
x = \pm \frac{\pi }{6} + k\pi
\end{array} \right.\)
LG c
\(\cos x\cos 2x = \cos 3x\)
Lời giải chi tiết:
\(\eqalign{ & \cos x\cos 2x = \cos 3x \cr & \Leftrightarrow {1 \over 2}\left( {\cos 3x + \cos x} \right) = \cos 3x \cr & \Leftrightarrow \cos 3x = \cos x \cr & \Leftrightarrow \left[ {\matrix{ {3x = x + k2\pi } \cr {3x = - x + k2\pi } \cr } } \right. \Leftrightarrow \left[ {\matrix{ {x = k\pi } \cr {x = k{\pi \over 2}} \cr } } \right.\cr& \Leftrightarrow x = k{\pi \over 2},k \in\mathbb Z \cr} \)
LG d
\(\tan 2x - \sin 2x + \cos 2x - 1 = 0\)
Lời giải chi tiết:
Điều kiện: \(\cos 2x \ne0\)
Ta có: \(\tan 2x = \dfrac{{\sin 2x}}{{\cos 2x}} \) \(\Rightarrow \sin 2x = \tan 2x\cos 2x\)
\(\eqalign{ & \tan 2x - \sin 2x + \cos 2x - 1 = 0 \cr & \Leftrightarrow \tan 2x - \tan 2x\cos 2x + \cos 2x - 1 = 0\cr & \Leftrightarrow \tan 2x\left( {1 - \cos 2x} \right) - \left( {1 - \cos 2x} \right) = 0 \cr & \Leftrightarrow \left( {1 - \cos 2x} \right)\left( {\tan 2x - 1} \right) = 0 \cr & \Leftrightarrow \left[ {\matrix{ {\tan 2x = 1} \cr {\cos 2x = 1} \cr } } \right. \cr} \)
\( \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{4} + k\pi \\
2x = k2\pi
\end{array} \right.\)
\( \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x = k\pi
\end{array} \right.,k \in Z\)
