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NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 8 2020
ĐKXĐ: ...
\(\Leftrightarrow\frac{cos\left(x+\frac{5\pi}{6}\right)}{cos\left(2x-\frac{\pi}{6}\right)}+\frac{sin\left(2x-\frac{\pi}{6}\right)}{cos\left(2x-\frac{\pi}{6}\right)}=0\)
\(\Leftrightarrow cos\left(x+\frac{5\pi}{6}\right)+sin\left(2x-\frac{\pi}{6}\right)=0\)
\(\Leftrightarrow cos\left(x+\frac{5\pi}{6}\right)=-sin\left(2x-\frac{\pi}{6}\right)\)
\(\Leftrightarrow cos\left(x+\frac{5\pi}{6}\right)=cos\left(2x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{3}=x+\frac{5\pi}{6}+k2\pi\\2x+\frac{\pi}{3}=-x-\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=-\frac{7\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 8 2020
\(\Leftrightarrow2sin\left(4x+\frac{6\pi}{5}\right)=\sqrt{3}\)
\(\Leftrightarrow sin\left(4x+\frac{6\pi}{5}\right)=\frac{\sqrt{3}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}4x+\frac{6\pi}{5}=\frac{\pi}{3}+k2\pi\\4x+\frac{6\pi}{5}=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{13\pi}{60}+\frac{k\pi}{2}\\x=-\frac{2\pi}{15}+\frac{k\pi}{2}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 8 2020
ĐKXĐ: ...
\(\Leftrightarrow sin\left(2x+\frac{3\pi}{4}\right)+cos\left(x+\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=-sin\left(2x+\frac{3\pi}{4}\right)\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=cos\left(2x+\frac{5\pi}{4}\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x+\frac{5\pi}{4}=x+\frac{\pi}{4}+k2\pi\\2x+\frac{5\pi}{4}=-x-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\pi+k2\pi\\x=-\frac{\pi}{2}+\frac{k2\pi}{3}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 8 2020
4.
\(\left\{{}\begin{matrix}cos^22x\ge0\\cos^23x\ge0\\cos^24x\ge0\end{matrix}\right.\) với mọi x
\(\Rightarrow cos^22x+cos^23x+cos^24x\ge0\) với mọi x
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}cos2x=0\\cos3x=0\\cos4x=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cos2x=0\\cos3x=0\\2cos^22x-1=0\end{matrix}\right.\)
Nếu \(cos2x=0\Rightarrow2cos^22x-1=-1\ne0\)
\(\Rightarrow\) Pt đã cho vô nghiệm
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 8 2020
3.
Ta có: \(\left\{{}\begin{matrix}cos^2x\ge0\\cos^22x\ge0\\cos^23x\ge0\end{matrix}\right.\) với mọi x
\(\Rightarrow cos^2x+cos^22x+cos^23x\ge0\) với mọi x
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}cosx=0\\cos2x=0\\cos3x=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx=0\\2cos^2x-1=0\\cos3x=0\end{matrix}\right.\)
Pt vô nghiệm (do nghiệm của pt thứ nhất ko thể là nghiệm của pt thứ 2)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 7 2019
a/ \(cos\left(3-2x\right)=-cosx=cos\left(\pi-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3-2x=\pi-x+k2\pi\\3-2x=x-\pi+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3-\pi+k2\pi\\x=1+\frac{\pi}{3}+\frac{k2\pi}{3}\end{matrix}\right.\)
b/ \(cos\left(3-2x\right)=sinx=cos\left(\frac{\pi}{2}-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}3-2x=\frac{\pi}{2}-x+k2\pi\\3-2x=x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3-\frac{\pi}{2}+k2\pi\\x=1+\frac{\pi}{6}+\frac{k2\pi}{3}\end{matrix}\right.\)
c/ Tương tự câu trên \(cos\left(3+2x\right)=-sinx=cos\left(\frac{\pi}{2}+x\right)\Rightarrow...\)
d/ \(sin4x-\sqrt{3}sin2x=0\)
\(\Leftrightarrow2sin2x.cos2x-\sqrt{3}sin2x=0\)
\(\Leftrightarrow sin2x\left(2cos2x-\sqrt{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=\frac{\sqrt{3}}{2}\end{matrix}\right.\) \(\Rightarrow...\)
HP
25 tháng 6 2021
a, \(cos^2x-cosx=0\)
\(\Leftrightarrow cosx\left(cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=0\end{matrix}\right.\)
b, \(2sin2x+\sqrt{2}sin4x=0\)
\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)
\(\Leftrightarrow sin2x\left(1+\sqrt{2}cos2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\1+\sqrt{2}cos2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\cos2x=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\2x=\dfrac{3\pi}{4}+k2\pi\\2x=\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\dfrac{3\pi}{8}+k\pi\\x=\dfrac{\pi}{8}+k\pi\end{matrix}\right.\)
25 tháng 6 2021
a, \(cos^2x-cosx=0\)
\(\Leftrightarrow cosx\left(cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=k2\pi\end{matrix}\right.\) (k ∈ Z)
Vậy...
b, \(2sin2x+\sqrt{2}sin4x=0\)
\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)
\(\Leftrightarrow2sin2x\left(1+\sqrt{2}cos2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=\dfrac{-\sqrt{2}}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\2x=\pm\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\pm\dfrac{3\pi}{8}+k\pi\end{matrix}\right.\)
Vậy...
c, \(8cos^2x+2sinx-7=0\)
\(\Leftrightarrow8\left(1-sin^2x\right)+2sinx-7=0\)
\(\Leftrightarrow8sin^2x-2sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-\dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\)
Vậy...
d, \(4cos^4x+cos^2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=\dfrac{3}{4}\\cos^2x=-1\left(loai\right)\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{cos2x+1}{2}=\dfrac{3}{4}\)
\(\Leftrightarrow cos2x=\dfrac{1}{2}\)
\(\Leftrightarrow2x=\pm\dfrac{\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+k\pi\)
Vậy...
e, \(\sqrt{3}tanx-6cotx+\left(2\sqrt{3}-3\right)=0\) (ĐK: \(x\ne\dfrac{k\pi}{2}\))
\(\Leftrightarrow\sqrt{3}tanx-\dfrac{6}{tanx}+\left(2\sqrt{3}-3\right)=0\)
\(\Leftrightarrow\sqrt{3}tan^2x+\left(2\sqrt{3}-3\right)tanx-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\\tanx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\left(tm\right)\\x=arctan\left(-2\right)+k\pi\end{matrix}\right.\)
Vậy...
28 tháng 6 2021
1.Pt \(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=sin\left(x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\left(\dfrac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\\2x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Rightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\)\(\left(k\in Z\right)\)
2.\(sin^22x+cos^23x=1\)
\(\Leftrightarrow\dfrac{1-cos4x}{2}+\dfrac{1+cos6x}{2}=1\)
\(\Leftrightarrow cos6x=cos4x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow x=\dfrac{k\pi}{5}\)\(\left(k\in Z\right)\) (Gộp nghiệm)
Vậy...
3. \(Pt\Leftrightarrow\left(sinx+sin3x\right)+\left(sin2x+sin4x\right)=0\)
\(\Leftrightarrow2.sin2x.cosx+2.sin3x.cosx=0\)
\(\Leftrightarrow2cosx\left(sin2x+sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=-sin2x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\sin3x=sin\left(\pi+2x\right)\end{matrix}\right.\)(\(k\in Z\))
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\)(\(k\in Z\))\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\) (\(k\in Z\))
Vậy...
4. Pt\(\Leftrightarrow\dfrac{1-cos2x}{2}+\dfrac{1-cos4x}{2}=\dfrac{1-cos6x}{2}\)
\(\Leftrightarrow cos2x+cos4x=1+cos6x\)
\(\Leftrightarrow2cos3x.cosx=2cos^23x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\cosx=cos3x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=-k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy...
