Giải các phương trình sau:
a) \(\sin^25x-1=0\)
b) \(\sin4x.\cos2x=0\)
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MA
1
30 tháng 6 2021
a)Đk:\(sinx\ne1\)
Pt\(\Leftrightarrow sin^2x+sinx=-2\left(sinx-1\right)\)
\(\Leftrightarrow sin^2x+3sinx-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{-3+\sqrt{17}}{2}\left(tm\right)\\sinx=\dfrac{-3-\sqrt{17}}{2}\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arcc.sin\left(\dfrac{-3+\sqrt{17}}{2}\right)+k2\pi\\x=\pi-arc.sin\left(\dfrac{-3+\sqrt{17}}{2}\right)+k2\pi\end{matrix}\right.\)(\(k\in Z\))
b)Đk:\(sinx\ne1\)
Pt \(\Leftrightarrow\dfrac{1-2sin^2x+sinx}{sinx-1}+1=0\)
\(\Leftrightarrow\dfrac{-\left(sinx-1\right)\left(2sinx+1\right)}{sinx-1}+1=0\)
\(\Leftrightarrow-\left(2sinx+1\right)+1=0\)
\(\Leftrightarrow sinx=0\) (tm)
\(\Leftrightarrow x=k\pi,k\in Z\)
Vậy...
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 9 2021
c.
\(\Leftrightarrow cos\left(x+12^0\right)+cos\left(90^0-78^0+x\right)=1\)
\(\Leftrightarrow2cos\left(x+12^0\right)=1\)
\(\Leftrightarrow cos\left(x+12^0\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+12^0=60^0+k360^0\\x+12^0=-60^0+k360^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=48^0+k360^0\\x=-72^0+k360^0\end{matrix}\right.\)
2.
Do \(-1\le sin\left(3x-27^0\right)\le1\) nên pt có nghiệm khi:
\(\left\{{}\begin{matrix}2m^2+m\ge-1\\2m^2+m\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m+1\ge0\left(luôn-đúng\right)\\2m^2+m-1\le0\end{matrix}\right.\)
\(\Rightarrow-1\le m\le\dfrac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 9 2021
a.
\(\Rightarrow\left[{}\begin{matrix}x+15^0=arccos\left(\dfrac{2}{5}\right)+k360^0\\x+15^0=-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-15^0+arccos\left(\dfrac{2}{5}\right)+k360^0\\x=-15^0-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)
b.
\(2x-10^0=arccot\left(4\right)+k180^0\)
\(\Rightarrow x=5^0+\dfrac{1}{2}arccot\left(4\right)+k90^0\)
QT
Quoc Tran Anh Le
Giáo viên
22 tháng 9 2023
a) Vì \(\sin \frac{\pi }{6} = \frac{1}{2}\) nên ta có phương trình \(sin2x = \sin \frac{\pi }{6}\)
\( \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{6} + k2\pi \\2x = \pi - \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{12}} + k\pi \\x = \frac{{5\pi }}{{12}} + k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)
\(\begin{array}{l}b,\,\,sin(x - \frac{\pi }{7}) = sin\frac{{2\pi }}{7}\\ \Leftrightarrow \left[ \begin{array}{l}x - \frac{\pi }{7} = \frac{{2\pi }}{7} + k2\pi \\x - \frac{\pi }{7} = \pi - \frac{{2\pi }}{7} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{3\pi }}{7} + k2\pi \\x = \frac{{6\pi }}{7} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)
\(\begin{array}{l}\;c)\;sin4x - cos\left( {x + \frac{\pi }{6}} \right) = 0\\ \Leftrightarrow sin4x = cos\left( {x + \frac{\pi }{6}} \right)\\ \Leftrightarrow sin4x = \sin \left( {\frac{\pi }{2} - x - \frac{\pi }{6}} \right)\\ \Leftrightarrow sin4x = \sin \left( {\frac{\pi }{3} - x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}4x = \frac{\pi }{3} - x + k2\pi \\4x = \pi - \frac{\pi }{3} + x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{15}} + k\frac{{2\pi }}{5}\\x = \frac{{2\pi }}{9} + k\frac{{2\pi }}{3}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)
QT
1
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a) \(\sin 2x + 1 - 2{\sin ^2}2x = 0\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\sin 2x = 1}\\{\sin 2x = - \frac{1}{2}}\end{array}\;\;\;} \right. \Leftrightarrow \;\left[ {\begin{array}{*{20}{c}}{\sin 2x = \sin \frac{\pi }{2}}\\{\sin 2x = \sin - \frac{\pi }{6}}\end{array}} \right.\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{2x = \frac{\pi }{2} + k2\pi }\\{2x = - \frac{\pi }{6} + k2\pi }\\{2x = \pi + \frac{\pi }{6} + k2\pi }\end{array}} \right.\;\;\)
\( \Leftrightarrow \;\left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{4} + k2\pi }\\{x = - \frac{\pi }{{12}} + k\pi }\\{x = \frac{{7\pi }}{{12}} + k\pi }\end{array}} \right.\;\;\left( {k \in \mathbb{Z}} \right)\)
b) \(\cos 3x = - \cos 7x\; \Leftrightarrow \cos 3x + \cos 7x = 0\;\; \Leftrightarrow 2\cos 5x\cos 2x = 0\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\cos 5x = 0}\\{\cos 2x = 0\;}\end{array}} \right.\;\;\)
\( \Leftrightarrow \left[ \begin{array}{l}\cos 5x = \cos \frac{\pi }{2}\\\cos 2x = \cos \frac{\pi }{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} + k2\pi \\5x = - \frac{\pi }{2} + k2\pi \\2x = \frac{\pi }{2} + k2\pi \\2x = - \frac{\pi }{2} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{10}} + \frac{{k2\pi }}{5}\\x = - \frac{\pi }{{10}} + \frac{{k2\pi }}{5}\\x = \frac{\pi }{4} + k\pi \\x = - \frac{\pi }{4} + k\pi \end{array} \right.;k \in Z\)
MA
1
HP
1 tháng 6 2021
\(4\left(sin^4x+cos^4x\right)+sin4x-2=0\)
\(\Leftrightarrow4\left(1-2sin^2x.cos^2x\right)+2sin2x.cos2x-2=0\)
\(\Leftrightarrow2-2sin^22x+2sin2x.cos2x=0\)
\(\Leftrightarrow2\left(1-sin^22x+sin2x.cos2x\right)=0\)
\(\Leftrightarrow2\left(cos^22x+sin2x.cos2x\right)=0\)
\(\Leftrightarrow2cos2x\left(cos2x+sin2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x+sin2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2};x=-\dfrac{\pi}{8}+\dfrac{k\pi}{2}\)
a)Pt\(\Leftrightarrow sin^25x=1\)
\(\Leftrightarrow\left[{}\begin{matrix}sin5x=1\\sin5x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{10}+\dfrac{k2\pi}{5}\\x=-\dfrac{\pi}{10}+\dfrac{k2\pi}{5}\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy...
b)Pt\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\cos2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2sin2x.cos2x=0\\cos2x=0\end{matrix}\right.\)\(\Rightarrow2.sin2x.cos2x=0\)\(\Leftrightarrow sin4x=0\Leftrightarrow x=\dfrac{k\pi}{4}\)\(\left(k\in Z\right)\)
Vậy...