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NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 8 2020
ĐKXĐ: \(sin3x\ne1\) \(\Rightarrow cos3x\ne0\)
\(\Rightarrow\) Phương trình vô nghiệm
15 tháng 11 2023
ĐKXĐ: \(\left\{{}\begin{matrix}sinx< >0\\sin2x< >0\\sin4x< >0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< >k\Omega\\2x< >k\Omega\\4x< >k\Omega\end{matrix}\right.\Leftrightarrow x\ne\dfrac{k\Omega}{4}\)
\(\dfrac{1}{sinx}+\dfrac{1}{sin2x}+\dfrac{1}{sin4x}=0\)
=>\(\dfrac{1}{sinx}+cotx+\dfrac{1}{sin2x}+cot2x+\dfrac{1}{sin4x}+cot4x=cotx+cot2x+cot4x\)
=>\(\dfrac{1+cosx}{sinx}+\dfrac{1+cos2x}{sin2x}+\dfrac{1+cos4x}{sin4x}=cotx+cot2x+cot4x\)
=>\(\dfrac{2\cdot cos^2\left(\dfrac{x}{2}\right)}{2\cdot sin\left(\dfrac{x}{2}\right)\cdot cos\left(\dfrac{x}{2}\right)}+\dfrac{2\cdot cos^2x}{2\cdot sinx\cdot cosx}+\dfrac{2\cdot cos^22x}{2\cdot sin2x\cdot cos2x}=cotx+cot2x+cot4x\)
=>\(\dfrac{cos\left(\dfrac{x}{2}\right)}{sin\left(\dfrac{x}{2}\right)}+\dfrac{cosx}{sinx}+\dfrac{cos2x}{sin2x}=cotx+cot2x+cot4x\)
=>\(cot\left(\dfrac{x}{2}\right)+cotx+cot2x=cotx+cot2x+cot4x\)
=>\(cot4x=cot\left(\dfrac{x}{2}\right)\)
=>\(\left\{{}\begin{matrix}4x=\dfrac{x}{2}+k\Omega\\4x< >k\Omega\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{7}{2}x=k\Omega\\x< >\dfrac{k\Omega}{4}\end{matrix}\right.\Leftrightarrow x=\dfrac{2}{7}k\Omega\)
MH
25 tháng 9 2023
\(\dfrac{1}{sinx}+\dfrac{1}{sin2x}+\dfrac{1}{sin4x}=0\)
\(\dfrac{1}{sinx}+cotx+\dfrac{1}{sin2x}+cot2x+\dfrac{1}{sin4x}+cot4x=cotx+cot2x+cot4x\)
\(\dfrac{1+cosx}{sinx}+\dfrac{1+cos2x}{sin2x}+\dfrac{1+cos4x}{sin4x}=cotx+cot2x+cot4x\)
\(\dfrac{2cos^2\dfrac{x}{2}}{2sin\dfrac{x}{2}.cos\dfrac{x}{2}}+\dfrac{2cos^2x}{2sinx.cosx}+\dfrac{2cos^22x}{2sin2x.cos2x}=cotx+cot2x+cot4x\)
\(\dfrac{cos\dfrac{x}{2}}{sin\dfrac{x}{2}}+\dfrac{cosx}{sinx}+\dfrac{cos2x}{sin2x}=cotx+cot2x+cot4x\)
\(cot\dfrac{x}{2}+cotx+cot2x=cotx+cot2x+cot4x\)
\(cot\dfrac{x}{2}=cot4x\)
\(\Rightarrow\dfrac{x}{2}=4x+k\text{π}\)
\(\Leftrightarrow x=-\dfrac{k2\text{π}}{7}\)
TV
0
NE
1
29 tháng 7 2020
\(\text{1) }3sinx-4cosx=1\\ \Leftrightarrow cos^2x+\left(\frac{4cosx+1}{3}\right)^2=1\\ \Leftrightarrow cosx=\frac{-4\pm6\sqrt{6}}{25}\\ \\ \Leftrightarrow x=arccos\left(\frac{-4\pm6\sqrt{6}}{25}\right)+k2\pi\)
\(2\text{) }\sqrt{3}sinx-cosx=1\\ \Leftrightarrow\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx=\frac{1}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sinx-sin\frac{\pi}{6}\cdot cosx=\frac{1}{2}\\ \Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=sin\frac{\pi}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=\frac{\pi}{6}+a2\pi\\x-\frac{\pi}{6}=\frac{5\pi}{6}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+a2\pi\\x=\pi+b2\pi\end{matrix}\right.\)
\(3\text{) }\sqrt{3}cosx+sinx=-2\\ \Leftrightarrow\frac{\sqrt{3}}{2}cosx+\frac{1}{2}sinx=-1\\ \Leftrightarrow sin\frac{\pi}{3}\cdot cosx+cos\frac{\pi}{3}\cdot sinx=-1\\ \Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=-1=sin\frac{3\pi}{2}\\ \\ \Leftrightarrow x+\frac{\pi}{3}=\frac{3\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{7\pi}{6}+k2\pi\)
\(4\text{) }cos4x-sin4x=1\\ \Leftrightarrow cos^24x+\left(cos4x-1\right)^2=1\\ \\ \Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+a\pi\\4x=b2\pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{a\pi}{4}\\x=\frac{b\pi}{2}\end{matrix}\right.\)
29 tháng 7 2020
\(5\text{) }\sqrt{3}cos4x+sin4x-2cos3x=0\\ \Leftrightarrow\frac{\sqrt{3}}{2}cos4x+\frac{1}{2}sin4x=cos3x\\ \Leftrightarrow cos\frac{\pi}{3}\cdot cos4x+sin\frac{\pi}{3}\cdot sin4x=cos3x\\ \Leftrightarrow cos\left(4x-\frac{\pi}{3}\right)=cos3x\\ \Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{3}=3x+a2\pi\\4x-\frac{\pi}{3}=-3x+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+a2\pi\\x=\frac{\pi}{21}+\frac{b2\pi}{7}\end{matrix}\right.\\ \Leftrightarrow x=\frac{\pi}{21}+\frac{k2\pi}{7}\)
\(6\text{) }cos^2x=3sin2x+3\\ \Leftrightarrow\frac{cos2x+1}{2}=3sin2x+3\)
Giải tương tự vd 1 và 4
7) Giải tương tự vd 1 và 4
H
9 tháng 9 2023
\(sin4x=-cos2x\\ \Leftrightarrow sin4x+cos2x=0\\ \Leftrightarrow2sin2x.cos2x+cos2x=0\\ \Leftrightarrow cos2x\left(2sin2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}cos2x=0\\2sin2x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{\pi}{2}+k\pi\\sin2x=-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\sin2x=sin\left(-\dfrac{\pi}{6}\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{12}+k\pi\\x=\dfrac{7\pi}{12}+k\pi\end{matrix}\right.\)
`HaNa♫D`
Đề là \(\sqrt{5}cos4x\) hay \(\sqrt{3}cos4x\) bạn?