Giải các phương trình sau:
LG a
\(\tan {x \over 2}\cos x - \sin 2x = 0\)
Phương pháp giải:
Hướng dẫn: Với điều kiện \(\cos {x \over 2} \ne 0,\) ta biến đổi phương trình đã cho thành
\(\cos x\left( {\tan {x \over 2} - 2\sin x} \right) = 0\) hay \(\cos x\sin {x \over 2}\left( {1 - 4{{\cos }^2}{x \over 2}} \right) = 0\)
Lời giải chi tiết:
ĐK: \(\cos {x \over 2} \ne 0\). Khi đó,
\(\tan {x \over 2}\cos x - \sin 2x = 0\)
\(\begin{array}{l}
\Leftrightarrow \tan \frac{x}{2}\cos x - 2\sin x\cos x = 0\\
\Leftrightarrow \cos x\left( {\tan \frac{x}{2} - 2\sin x} \right) = 0\\
\Leftrightarrow \cos x.\left( {\frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}} - 4\sin \frac{x}{2}\cos \frac{x}{2}} \right) = 0\\
\Leftrightarrow \cos x\sin \frac{x}{2}\left( {\frac{1}{{\cos \frac{x}{2}}} - 4\cos \frac{x}{2}} \right) = 0\\
\Leftrightarrow \cos x\sin \frac{x}{2}\left( {1 - 4{{\cos }^2}\frac{x}{2}} \right) = 0\\
\Leftrightarrow \cos x\sin \frac{x}{2}\left( {1 - 4.\frac{{1 + \cos x}}{2}} \right) = 0\\
\Leftrightarrow \cos x\sin \frac{x}{2}\left( { - 1 - 2\cos x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin \frac{x}{2} = 0\\
\cos x = - \frac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{2} + k\pi \\
x = k2\pi \\
x = \pm \frac{{2\pi }}{3} + k2\pi
\end{array} \right.
\end{array}\)
Vậy \(x = \pm {{2\pi } \over 3} + 2k\pi ,x = 2k\pi ,x = {\pi \over 2} + k\pi \)
LG b
\({\sin ^6}x + 3{\sin ^2}x\cos 4x + {\cos ^6}x = 1\)
Phương pháp giải:
Hướng dẫn: Áp dụng hằng đẳng thức dễ thấy:
\({a^6} + {b^6} = {\left( {{a^2} + {b^2}} \right)^3} - 3{a^2}{b^2}\left( {{a^2} + {b^2}} \right)\)
Lời giải chi tiết:
\(x = {{k\pi } \over 2}\)
Ta có:
\(\eqalign{
& {\sin ^6}x + 3{\sin ^2}x\cos 4x + {\cos ^6}x = 1 \cr
& \Leftrightarrow {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^3} \cr&- 3{\sin ^2}x{\cos ^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\cr& + 3{\sin ^2}x\cos x = 1 \cr
& \Leftrightarrow - 3{\sin ^2}x{\cos ^2}x + 3{\sin ^2}x\cos x = 0 \cr} \)
\(\begin{array}{l}
\Leftrightarrow 3{\sin ^2}x\cos x\left( {1 - \cos x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\cos x = 0\\
\cos x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \frac{\pi }{2} + k\pi \\
x = k2\pi
\end{array} \right.\\
\Leftrightarrow x = \frac{{k\pi }}{2}
\end{array}\)
LG c
\({\sin ^3}x\cos x - \sin x{\cos ^3}x = {{\sqrt 2 } \over 8}\)
Phương pháp giải:
Hướng dẫn: \({\sin ^3}x\cos x - \sin x{\cos ^3}x = {{\sqrt 2 } \over 8}\)
\(\Leftrightarrow \sin x\cos x\left( {{{\sin }^2}x-{{\cos }^2}x} \right) = {{\sqrt 2 } \over 8}\)
\( \Leftrightarrow {1 \over 2}\sin 2x\cos 2x = - {{\sqrt 2 } \over 8}\)
Lời giải chi tiết:
\({\sin ^3}x\cos x - \sin x{\cos ^3}x = {{\sqrt 2 } \over 8}\)
\(\Leftrightarrow \sin x\cos x\left( {{{\sin }^2}x-{{\cos }^2}x} \right) = {{\sqrt 2 } \over 8}\)
\( \Leftrightarrow {1 \over 2}\sin 2x\cos 2x = - {{\sqrt 2 } \over 8}\)
\(\begin{array}{l}
\Leftrightarrow \frac{1}{4}\sin 4x = - \frac{{\sqrt 2 }}{8}\\
\Leftrightarrow \sin 4x = - \frac{{\sqrt 2 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
4x = - \frac{\pi }{4} + k2\pi \\
4x = \frac{{5\pi }}{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \frac{\pi }{{16}} + \frac{{k\pi }}{2}\\
x = \frac{{5\pi }}{{16}} + \frac{{k\pi }}{2}
\end{array} \right.
\end{array}\)
Vậy \(x = - {\pi \over 16} + {{k\pi } \over 2},x = {{5\pi } \over {16}} + {{k\pi } \over 2}\)
LG d
\({\sin ^2}x + \sin x\cos 4x + {\cos ^2}4x = {3 \over 4}\)
Lời giải chi tiết:
Ta có:
\(\eqalign{
& {\sin ^2}x + \sin x\cos 4x + {{{{\cos }^2}4x} \over 4} + {3 \over 4}{\cos ^2}4x = {3 \over 4} \cr
& \Leftrightarrow {\left( {\sin x + {1 \over 2}\cos 4x} \right)^2} = {3 \over 4}\left( {1 - {{\cos }^2}4x} \right)\cr& \Leftrightarrow {\left( {\sin x + {1 \over 2}\cos 4x} \right)^2} = {3 \over 4}{\sin ^2}4x \cr
& \Leftrightarrow \left[ \matrix{
\sin x + {1 \over 2}\cos 4x = {{\sqrt 3 } \over 2}\sin 4x \hfill \cr
\sin x + {1 \over 2}\cos 4x = - {{\sqrt 3 } \over 2}\sin 4x \hfill \cr} \right. \cr& \Leftrightarrow \left[ \matrix{
\cos {\pi \over 6}\sin 4x - \sin {\pi \over 6}\cos 4x = \sin x \hfill \cr
\sin {\pi \over 6}\cos 4x + \cos {\pi \over 6}\sin 4x = \sin \left( { - x} \right) \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
\sin \left( {4x - {\pi \over 6}} \right) = \sin x \hfill \cr
\sin \left( {4x + {\pi \over 6}} \right) = \sin \left( { - x} \right) \hfill \cr} \right. \cr} \)
\(\begin{array}{l}
\Leftrightarrow \left[ \begin{array}{l}
4x - \frac{\pi }{6} = x + k2\pi \\
4x - \frac{\pi }{6} = \pi - x + k2\pi \\
4x + \frac{\pi }{6} = - x + k2\pi \\
4x + \frac{\pi }{6} = \pi + x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{{18}} + \frac{{k2\pi }}{3}\\
x = \frac{{7\pi }}{{30}} + \frac{{k2\pi }}{5}\\
x = - \frac{\pi }{{30}} + \frac{{k2\pi }}{5}\\
x = \frac{{5\pi }}{{18}} + \frac{{k2\pi }}{3}
\end{array} \right.
\end{array}\)
Vậy \(x = {\pi \over {18}} + k{{2\pi } \over 3},x = {{7\pi } \over {30}} + k{{2\pi } \over 5},\) \(x = - {\pi \over {30}} + k{{2\pi } \over 5},x = {{5\pi } \over {18}} + k{{2\pi } \over 3}\).
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