Giải phương trình sau:
1) \(\frac{1}{cosx}+\frac{1}{sin2x}=\frac{2}{sin4x}\)
2) \(\frac{sin^4x+cos^4x}{5sin2x}=\frac{1}{2}cot2x-\frac{1}{8sin2x}\)
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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 \(sin2x = \sin \frac{\pi }{6}\)
\( \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{6} + k2\pi \\2x = \pi - \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{12}} + k\pi \\x = \frac{{5\pi }}{{12}} + k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)
\(\begin{array}{l}b,\,\,sin(x - \frac{\pi }{7}) = sin\frac{{2\pi }}{7}\\ \Leftrightarrow \left[ \begin{array}{l}x - \frac{\pi }{7} = \frac{{2\pi }}{7} + k2\pi \\x - \frac{\pi }{7} = \pi - \frac{{2\pi }}{7} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{3\pi }}{7} + k2\pi \\x = \frac{{6\pi }}{7} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)
\(\begin{array}{l}\;c)\;sin4x - cos\left( {x + \frac{\pi }{6}} \right) = 0\\ \Leftrightarrow sin4x = cos\left( {x + \frac{\pi }{6}} \right)\\ \Leftrightarrow sin4x = \sin \left( {\frac{\pi }{2} - x - \frac{\pi }{6}} \right)\\ \Leftrightarrow sin4x = \sin \left( {\frac{\pi }{3} - x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}4x = \frac{\pi }{3} - x + k2\pi \\4x = \pi - \frac{\pi }{3} + x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{15}} + k\frac{{2\pi }}{5}\\x = \frac{{2\pi }}{9} + k\frac{{2\pi }}{3}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 8 2020
a/
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x-2\left(1-sin^22x\right)=0\)
\(\Leftrightarrow1-\frac{1}{2}\left(cos6x+cos2x\right)-2cos^22x=0\)
\(\Leftrightarrow1-cos4x.cos2x-2cos^22x=0\)
\(\Leftrightarrow2cos^22x-1+cos4x.cos2x=0\)
\(\Leftrightarrow cos4x+cos4x.cos2x=0\)
\(\Leftrightarrow cos4x\left(cos2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\2x=\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{2}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 8 2020
d/
ĐKXĐ: \(sin2x\ne0\) \(\Leftrightarrow2x\ne k\pi\)
\(\Leftrightarrow1+\frac{cos2x}{sin2x}=\frac{1-cos2x}{sin^22x}\)
\(\Leftrightarrow sin^22x+sin2x.cos2x=1-cos2x\)
\(\Leftrightarrow sin^22x-1+sin2x.cos2x+cos2x=0\)
\(\Leftrightarrow-cos^22x+sin2x.cos2x+cos2x=0\)
\(\Leftrightarrow cos2x\left(sin2x-cos2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin2x-cos2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin\left(2x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\2x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\2x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=k\pi\left(l\right)\\x=\frac{3\pi}{4}+k\pi\end{matrix}\right.\)
MA
0
29 tháng 9 2020
a.\(\frac{k\Pi}{2}+\frac{\alpha}{2}\)
b.\(\left\{{}\begin{matrix}x=\frac{1}{4}arcsin\left(\frac{1}{3}\right)+\frac{k\Pi}{2}-\frac{1}{8}\\x=\Pi-\frac{1}{4}arcsin\left(\frac{1}{3}\right)+\frac{k\Pi}{2}-\frac{1}{8}\end{matrix}\right.\)
Câu 1:
ĐKXĐ: \(sin4x\ne0\Rightarrow x\ne\frac{k\pi}{4}\)
\(\Leftrightarrow\frac{sin4x}{cosx}+\frac{sin4x}{sin2x}=2\Leftrightarrow4sinx.cos2x+2cos2x=2\)
\(\Leftrightarrow cos2x\left(2sinx+1\right)=1\Leftrightarrow\left(1-2sin^2x\right)\cdot\left(2sinx+1\right)=1\)
\(\Leftrightarrow2sinx-4sin^3x-2sin^2x=0\)
\(\Leftrightarrow sinx\left(-2sin^2x-sinx+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=0\left(l\right)\\sinx=-1\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
b/
ĐKXĐ: \(sin2x\ne0\Rightarrow x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow\frac{sin^4x+cos^4x}{5}=\frac{1}{2}cos2x-\frac{1}{8}\)
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2\left(sinx.cosx\right)^2=\frac{5}{2}cos2x-\frac{5}{8}\)
\(\Leftrightarrow1-\frac{1}{2}sin^22x=\frac{5}{2}cos2x-\frac{5}{8}\)
\(\Leftrightarrow1-\frac{1}{2}\left(1-cos^22x\right)=\frac{5}{2}cos2x-\frac{5}{8}\)
\(\Leftrightarrow\frac{1}{2}cos^22x-\frac{5}{2}cos2x+\frac{9}{8}=0\)
\(\Rightarrow\left[{}\begin{matrix}cos2x=\frac{9}{2}>1\left(l\right)\\cos2x=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow x=\pm\frac{\pi}{6}+k\pi\)