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HP
17 tháng 8 2021
\(cos\left(4x+60^o\right)-5cos\left(2x+30^o\right)+4=0\)
\(\Leftrightarrow2cos^2\left(2x+30^o\right)-5cos\left(2x+30^o\right)+3=0\)
\(\Leftrightarrow\left[cos\left(2x+30^o\right)-1\right]\left[2cos\left(2x+30^o\right)-3\right]=0\)
\(\Leftrightarrow cos\left(2x+30^o\right)=1\)
\(\Leftrightarrow2x+30^o=k.360^o\)
\(\Leftrightarrow x=-15^o+k.180^o\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 8 2020
ĐKXĐ: \(cos\left(x-30^0\right)\ne0\Leftrightarrow x\ne120^0+k180^0\)
\(tan\left(x-30^0\right)cos\left(2x-150^0\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tan\left(x-30^0\right)=0\\cos\left(2x-150^0\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-30^0=k180^0\\2x-150^0=90^0+k180^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=30^0+k180^0\\x=120^0+k90^0\end{matrix}\right.\)
\(\Rightarrow x=30^0+k180^0\)
NT
1
HT
21 tháng 8 2020
a/ \(\Leftrightarrow\cos\left(\frac{\pi}{7}-3x\right)=\cos\left(-\frac{5}{6}\pi\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{\pi}{7}-3x=-\frac{5}{6}\pi+k2\pi\\\frac{\pi}{7}-3x=\frac{5}{6}\pi+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{41}{126}\pi-\frac{2}{3}k\pi\\x=-\frac{29}{42}\pi-\frac{2}{3}k\pi\end{matrix}\right.\)
b/ \(\Leftrightarrow\sin\left(90^0-\frac{x}{3}\right)=\sin\left(2x+30^0\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}90^0-\frac{x}{3}=2x+30^0+k180^0\\90^0-\frac{x}{3}=180^0-2x-30^0+k180^0\end{matrix}\right.\Leftrightarrow...\)
c/ \(DKXD:\cos\left(30^0-2x\right)\ne0\Leftrightarrow30^0-2x\ne90^0+k180^0\Leftrightarrow x\ne-30^0-k90^0\)
\(\Leftrightarrow30^0-2x=60^0+k180^0\Leftrightarrow x=-15^0-k90^0\left(tm\right)\)
d/ \(DKXD:\sin\left(30^0-2x\right)\ne0\Leftrightarrow30^0-2x\ne k180^0\Leftrightarrow x\ne15^0-k90^0\)
\(\Leftrightarrow30^0-2x=30^0+k.180^0\Leftrightarrow x=-k.90^0\left(tm\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 9 2019
Đặt \(x-30^0=a\)
\(\sqrt{2}sin2a-2sina=\sqrt{2}cos2a\)
\(\Leftrightarrow sin2a-cos2a=\sqrt{2}sina\)
\(\Leftrightarrow\sqrt{2}sin\left(2a-45^0\right)=\sqrt{2}sina\)
\(\Leftrightarrow sin\left(2a-45^0\right)=sina\)
\(\Leftrightarrow sin\left(2x-105^0\right)=sin\left(x-30^0\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x-105^0=x-30^0+k360^0\\2x-105^0=180^0-x+30^0+k360^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=75^0+k.360^0\\x=105^0+k.120^0\end{matrix}\right.\)
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. \(Pt\Leftrightarrow\left(sinx+sin3x\right)+\left(sin2x+sin4x\right)=0\)
\(\Leftrightarrow2.sin2x.cosx+2.sin3x.cosx=0\)
\(\Leftrightarrow2cosx\left(sin2x+sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=-sin2x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\sin3x=sin\left(\pi+2x\right)\end{matrix}\right.\)(\(k\in Z\))
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\)(\(k\in Z\))\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\) (\(k\in Z\))
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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...
