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NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 8 2020
c/
\(\Leftrightarrow cos3x-\sqrt{3}sin3x=\sqrt{3}cos2x-sin2x\)
\(\Leftrightarrow\frac{1}{2}cos3x-\frac{\sqrt{3}}{2}sin3x=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow cos\left(3x+\frac{\pi}{3}\right)=cos\left(2x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{3}=2x+\frac{\pi}{6}+k2\pi\\3x+\frac{\pi}{3}=-2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=-\frac{\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 8 2020
b/
\(\Leftrightarrow cosx-\sqrt{3}sinx=sin2x-\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=sin\left(2x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=sin\left(\frac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{5\pi}{6}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 7 2020
\(sin3x-sinx+sin2x=0\)
\(\Leftrightarrow2cos2x.sinx+2sinx.cosx=0\)
\(\Leftrightarrow sinx\left(cos2x+cosx\right)=0\)
\(\Leftrightarrow2sinx.cos\frac{3x}{2}.cos\frac{x}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=0\\cos\frac{x}{2}=0\\cos\frac{3x}{2}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\\frac{x}{2}=\frac{\pi}{2}+k\pi\\\frac{3x}{2}=\frac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\pi+k2\pi\\x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 7 2020
\(cosx+cos3x+cos2x+cos4x=0\)
\(\Leftrightarrow2cos2x.cosx+2cos3x.cosx=0\)
\(\Leftrightarrow cosx\left(cos2x+cos3x\right)=0\)
\(\Leftrightarrow2cosx.cos\frac{5x}{2}.cos\frac{x}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cos\frac{x}{2}=0\\cos\frac{5x}{2}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\\frac{x}{2}=\frac{\pi}{2}+k\pi\\\frac{5x}{2}=\frac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\frac{\pi}{5}+\frac{k2\pi}{5}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 8 2020
\(\Leftrightarrow sin4x\left(sin5x+sin3x\right)-sin2x.sinx=0\)
\(\Leftrightarrow2sin^24x.cosx-2sin^2x.cosx=0\)
\(\Leftrightarrow cosx\left(2sin^24x-2sin^2x\right)=0\)
\(\Leftrightarrow cosx\left(1-cos8x-1+cos2x\right)=0\)
\(\Leftrightarrow cosx\left(cos2x-cos8x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos8x=cos2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\8x=2x+k2\pi\\8x=-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{k\pi}{3}\\x=\frac{k\pi}{5}\end{matrix}\right.\)
21 tháng 7 2022
\(\Leftrightarrow\sin3x+\sin x+\sin2x=0\)
\(\Leftrightarrow2\cdot\sin2x\cdot\cos x+\sin2x=0\)
\(\Leftrightarrow\sin2x\left(2\cos x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\Pi\\\cos x=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\Pi}{2}\\x=\dfrac{2}{3}\Pi+k2\Pi\\x=-\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)
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
\(\dfrac{1}{sinx}+\dfrac{1}{sin2x}+\dfrac{1}{sin4x}=0\)
\(\dfrac{1}{sinx}+cotx+\dfrac{1}{sin2x}+cot2x+\dfrac{1}{sin4x}+cot4x=cotx+cot2x+cot4x\)
\(\dfrac{1+cosx}{sinx}+\dfrac{1+cos2x}{sin2x}+\dfrac{1+cos4x}{sin4x}=cotx+cot2x+cot4x\)
\(\dfrac{2cos^2\dfrac{x}{2}}{2sin\dfrac{x}{2}.cos\dfrac{x}{2}}+\dfrac{2cos^2x}{2sinx.cosx}+\dfrac{2cos^22x}{2sin2x.cos2x}=cotx+cot2x+cot4x\)
\(\dfrac{cos\dfrac{x}{2}}{sin\dfrac{x}{2}}+\dfrac{cosx}{sinx}+\dfrac{cos2x}{sin2x}=cotx+cot2x+cot4x\)
\(cot\dfrac{x}{2}+cotx+cot2x=cotx+cot2x+cot4x\)
\(cot\dfrac{x}{2}=cot4x\)
\(\Rightarrow\dfrac{x}{2}=4x+k\text{π}\)
\(\Leftrightarrow x=-\dfrac{k2\text{π}}{7}\)
21 tháng 7 2022
b: \(\Leftrightarrow2\cdot\cos2x\cdot\cos x+2\cdot\sin x\cdot\cos2x=\sqrt{2}\cdot\cos2x\)
\(\Leftrightarrow2\cdot\cos2x\left(\sin x+\cos x\right)=\sqrt{2}\cdot\cos2x\)
\(\Leftrightarrow\sqrt{2}\cdot\cos2x\cdot\left[\sqrt{2}\cdot\sqrt{2}\cdot\sin\left(x+\dfrac{\Pi}{4}\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\cos2x=0\\\sin\left(x+\dfrac{\Pi}{4}\right)=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\Pi}{2}+k\Pi\\x+\dfrac{\Pi}{4}=\dfrac{\Pi}{6}+k2\Pi\\x+\dfrac{\Pi}{4}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\)
\(\Leftrightarrow x\in\left\{\dfrac{\Pi}{4}+\dfrac{k\Pi}{2};\dfrac{-1}{12}\Pi+k2\Pi;\dfrac{7}{12}\Pi+k2\Pi\right\}\)
c: \(\Leftrightarrow2\cdot\sin2x\cdot\cos x+\sin2x=2\cdot\cos2x\cdot\cos x+\cos2x\)
\(\Leftrightarrow\sin2x\left(2\cos x+1\right)=\cos2x\left(2\cos x+1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\sin2x=\cos2x=\sin\left(\dfrac{\Pi}{2}-2x\right)\\\cos x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{4}\\\\x=-\dfrac{2}{3}\Pi+k2\Pi\\x=\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)
\(\Leftrightarrow\sin x+2\sin x\cos x+3\sin x-4\sin^3x+2\sin2x\cos2x=0\)
\(\Leftrightarrow4\sin x-4\sin^3x+2\sin x\cos x+4\sin x\cos x\left(1-2\sin^2x\right)=0\)
\(\Leftrightarrow4\sin x\left(1-\sin^2x\right)+2\sin x\cos x+4\sin x\cos x-8\sin^3x\cos x=0\)
\(\Leftrightarrow4\sin x\cos^2x+6\sin x\cos x-8\sin^3x\cos x=0\)
\(\Leftrightarrow2\sin x\cos x\left(2\cos x+3+4\sin^2x\right)=0\)
\(\Leftrightarrow\sin2x\left(2\cos x+3+4-4\cos^2x\right)=0\)
Đến đây là phương trình tích. Trong ngoặc đơn là PT bậc 2 đối với cosx. Bạn tự làm nốt nhé