2.cos3x.cosx = cos5x + cos2x+4
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HP
23 tháng 8 2021
2.
\(sin3x+cos2x=1+2sinx.cos2x\)
\(\Leftrightarrow sin3x+cos2x=1+sin3x-sinx\)
\(\Leftrightarrow cos2x+sinx-1=0\)
\(\Leftrightarrow-2sin^2x+sinx=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
HP
23 tháng 8 2021
1.
\(cos3x-cos4x+cos5x=0\)
\(\Leftrightarrow cos3x+cos5x-cos4x=0\)
\(\Leftrightarrow2cos4x.cosx-cos4x=0\)
\(\Leftrightarrow\left(2cosx-1\right)cos4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{1}{2}\\cos4x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\4x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\end{matrix}\right.\)
NT
1
21 tháng 7 2022
a: \(\Leftrightarrow\dfrac{1}{2}\cdot\cos2x\cdot\cos x-\cos2x=0\)
\(\Leftrightarrow\cos2x=0\)
\(\Leftrightarrow2x=\dfrac{\Pi}{2}+k\Pi\)
hay \(x=\dfrac{\Pi}{4}+\dfrac{k\Pi}{2}\)
b: \(\Leftrightarrow\dfrac{1}{2}\cdot\left[\cos\left(5x-x\right)-\cos\left(5x+x\right)\right]=\dfrac{1}{2}\cdot\left[\cos\left(3x-2x\right)-\cos5x\right]\)
\(\Leftrightarrow\cos4x-\cos6x=\cos x-\cos5x\)
\(\Leftrightarrow x=\dfrac{\Pi}{2}+k\Pi\)
HP
12 tháng 7 2021
1.
\(2cos4x-3=0\)
\(\Leftrightarrow cos4x=\dfrac{3}{2}\)
Mà \(cos4x\in\left[-1;1\right]\)
\(\Rightarrow\) phương trình vô nghiệm.
2.
\(cos5x+2=0\)
\(\Leftrightarrow cos5x=-2\)
Mà \(cos5x\in\left[-1;1\right]\)
\(\Rightarrow\) phương trình vô nghiệm.
HP
12 tháng 7 2021
3.
\(cos2x+0,7=0\)
\(\Leftrightarrow cos2x=-\dfrac{7}{10}\)
\(\Leftrightarrow2x=\pm arccos\left(-\dfrac{7}{10}\right)+k2\pi\)
\(\Leftrightarrow x=\pm\dfrac{arccos\left(-\dfrac{7}{10}\right)}{2}+k\pi\)
4.
\(cos^22x-\dfrac{1}{4}=0\)
\(\Leftrightarrow cos^22x=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-\dfrac{1}{2}\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\pm\dfrac{2\pi}{3}+k2\pi\\2x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k\pi\\x=\pm\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
1 tháng 1 2021
\(cos5\text{x}+sin^2x=cos^2x\\ \Leftrightarrow cos5\text{x}=cos2x\\ \\ \Leftrightarrow\left[{}\begin{matrix}5x=2x+k2\pi\\5x=-2x+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}xx=\dfrac{k2\pi}{3}\\x\text{ }=\dfrac{k2\pi}{7}\end{matrix}\right.\)
NN
0
8 tháng 6 2022
\(\cos\left(2x\right)-\sin\left(3x\right)+\cos\left(5x\right)=\sin\left(10x\right)+\cos\left(8x\right)\)
\(\Leftrightarrow\cos\left(2x\right)-\cos\left(8x\right)+\cos\left(5x\right)-\sin\left(3x\right)-\sin\left(10x\right)=0\)
\(\Leftrightarrow-\left(\cos\left(8x\right)-\cos\left(2x\right)\right)+\cos\left(5x\right)-\left(\sin(10x\right)+\sin\left(3x\right))=0\)
\(\Leftrightarrow2\sin\left(5x\right)\sin\left(3x\right)+\cos\left(5x\right)-\sin\left(3x\right)-2\sin\left(5x\right)\cos\left(5x\right)=0\)
\(\Leftrightarrow2\sin\left(5x\right)(\sin\left(3x\right)-cos\left(5x\right))-\left(sin\left(3x\right)-cos\left(5x\right)\right)=0\)
\(\Leftrightarrow\left(2sin\left(5x\right)-1\right)\left(sin\left(3x\right)-cos\left(5x\right)\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(5x\right)=\dfrac{1}{2}\\sin\left(3x\right)=cos\left(5x\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin\left(5x\right)=\dfrac{1}{2}\\sin\left(3x\right)=sin\left(\dfrac{\pi}{2}-5x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{\pi}{30}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{5}\end{matrix}\right.\\\left[{}\begin{matrix}3x=\dfrac{\pi}{2}-5x+k2\pi\\3x=\pi-\dfrac{\pi}{2}+5x+k2\pi\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{30}+\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{16}+\dfrac{k\pi}{4}\\x=-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 8 2020
ĐKXĐ: \(x\ne\frac{\pi}{6}+\frac{k\pi}{3}\)
\(\Leftrightarrow\frac{cos^2x-cos3x.cos5x}{cos3x.cosx}-4\left[1-2sin^2\left(2x+\frac{11\pi}{2}\right)\right]-4cos2x=0\)
\(\Leftrightarrow\frac{2cos^2x-cos2x-cos8x}{cos4x+cos2x}-4cos\left(4x+11\pi\right)-4cos2x=0\)
\(\Leftrightarrow\frac{1-cos8x}{cos4x+cos2x}+4cos4x-4cos2x=0\)
\(\Leftrightarrow1-cos8x+4\left(cos4x-cos2x\right)\left(cos4x+cos2x\right)=0\)
\(\Leftrightarrow1-cos8x+4cos^24x-4cos^22x=0\)
\(\Leftrightarrow1-\left(2cos^24x-1\right)+4cos^24x-2\left(1+cos4x\right)=0\)
\(\Leftrightarrow cos^24x-cos4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos4x=1\end{matrix}\right.\) \(\Leftrightarrow...\)