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NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2020
c/
\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=cos3x\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=cos3x\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=3x+k2\pi\\x+\frac{\pi}{3}=-3x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=\frac{\pi}{12}+\frac{k\pi}{2}\end{matrix}\right.\)
d/
\(\Leftrightarrow\frac{1}{2}sin3x-\frac{\sqrt{3}}{2}cos3x=sin2x\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{3}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{3}=2x+k2\pi\\3x-\frac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=\frac{4\pi}{15}+\frac{k2\pi}{5}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2020
a/
\(\Leftrightarrow\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx=sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{6}\right)\)
\(\Rightarrow x+\frac{\pi}{3}=\pi-x-\frac{\pi}{6}+k2\pi\)
\(\Rightarrow x=\frac{\pi}{4}+k\pi\)
b/
\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx=sin\frac{\pi}{12}\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{6}\right)=sin\frac{\pi}{12}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=\frac{\pi}{12}+k2\pi\\x+\frac{\pi}{6}=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+k2\pi\\x=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 10 2020
d.
\(\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+k2\pi\)
e.
\(\Leftrightarrow cosx.cos\left(\frac{\pi}{12}\right)-sinx.sin\left(\frac{\pi}{12}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{12}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{12}=\frac{\pi}{3}+k2\pi\\x+\frac{\pi}{12}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 10 2020
2.a.
ĐKXĐ: ...
\(\sqrt{3}tanx-\frac{6}{tanx}+2\sqrt{3}-3=0\)
\(\Leftrightarrow\sqrt{3}tan^2x+\left(2\sqrt{3}-3\right)tanx-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-2\\tanx=\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arctan\left(-2\right)+k\pi\\x=\frac{\pi}{3}+k\pi\end{matrix}\right.\)
b.
ĐKXĐ: \(x\ne k\pi\)
\(1-sin2x=2sin^2x\)
\(\Leftrightarrow1-2sin^2x-sin2x=0\)
\(\Leftrightarrow cos2x-sin2x=0\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 10 2020
ĐKXĐ: ..
\(\frac{sin3x+sinx+sin2x}{cos3x+cosx+cos2x}=\sqrt{3}\)
\(\Leftrightarrow\frac{2sin2x.cosx+sin2x}{2cos2x.cosx+cos2x}=\sqrt{3}\)
\(\Leftrightarrow\frac{sin2x\left(2cosx+1\right)}{cos2x\left(2cosx+1\right)}=\sqrt{3}\)
\(\Leftrightarrow tan2x=\sqrt{3}\)
\(\Leftrightarrow x=\frac{\pi}{6}+\frac{k\pi}{2}\)
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: \(\Leftrightarrow2\cdot\sin2x\cdot\cos x+\sin2x=2\cdot\cos2x\cdot\cos x+\cos2x\)
\(\Leftrightarrow\sin2x\left(2\cos x+1\right)=\cos2x\left(2\cos x+1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\sin2x=\cos2x=\sin\left(\dfrac{\Pi}{2}-2x\right)\\\cos x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{4}\\\\x=-\dfrac{2}{3}\Pi+k2\Pi\\x=\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2020
c/
\(\Leftrightarrow\sqrt{3}sin3x-cos3x=sin2x-\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin3x-\frac{1}{2}cos3x=\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{6}\right)=sin\left(2x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{6}=2x-\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{6}=\pi-2x+\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2020
e/
\(\Leftrightarrow\frac{1}{2}sin8x-\frac{\sqrt{3}}{2}cos8x=\frac{\sqrt{3}}{2}sin6x+\frac{1}{2}cos6x\)
\(\Leftrightarrow sin\left(8x-\frac{\pi}{3}\right)=sin\left(6x+\frac{\pi}{6}\right)\)
\(\Rightarrow\left[{}\begin{matrix}8x-\frac{\pi}{3}=6x+\frac{\pi}{6}+k2\pi\\8x-\frac{\pi}{3}=\pi-6x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{28}+\frac{k\pi}{7}\end{matrix}\right.\)
NT
1
12 tháng 7 2016
<=> 2.cos2x.cosx = \(2\sqrt{3}\) cos2x.sinx
<=> cos2x.cosx = \(\sqrt{3}\) cos2x.sinx
<=> cos2x.( cosx - \(\sqrt{3}\) sinx) = 0
<=> \(\left[\begin{array}{nghiempt}cos2x=0\\cosx-\sqrt{3}sinx=0\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}2x=\frac{\pi}{2}+k\pi\\\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=0
\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=\frac{\pi}{4}+\frac{k\pi}{2}\\sin\frac{\pi}{6}.cosx-cos\frac{\pi}{6}.sinx=0\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=\frac{\pi}{4}+\frac{k\pi}{2}\\sin\left(\frac{\pi}{6}-x\right)=0\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{6}-k\pi\end{array}\right.\) (k\(\in\)Z)
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 7 2020
c/
\(\Leftrightarrow\sqrt{2}sin\left(3x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{\sqrt{2}}\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}3x-\frac{\pi}{4}=\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{4}=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{7\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{36}+\frac{k2\pi}{3}\end{matrix}\right.\)
d/
\(\Leftrightarrow2sinx.cosx+1-2sin^2x=1\)
\(\Leftrightarrow2sinx\left(cosx-sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=cosx\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 7 2020
a/
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin5x-\frac{1}{2}cos5x=-1\)
\(\Leftrightarrow sin\left(5x-\frac{\pi}{6}\right)=-1\)
\(\Leftrightarrow5x-\frac{\pi}{6}=-\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=-\frac{\pi}{15}+\frac{k2\pi}{5}\)
b/
\(\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 10 2020
1.
\(\Leftrightarrow sin^2x\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cos^2x\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(1+cosx\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(sinx+cosx+sinx.cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\Leftrightarrow...\\sinx+cosx+sinx.cosx-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
Đặt \(sinx+cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow t+\frac{t^2-1}{2}-1=0\)
\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 10 2020
2.
\(\Leftrightarrow\sqrt{3}sinx.cosx+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)
\(\Leftrightarrow cosx\left(\sqrt{3}sinx+\sqrt{2}cosx+\sqrt{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Leftrightarrow...\\\sqrt{3}sinx+\sqrt{2}cosx=-\sqrt{6}\left(1\right)\end{matrix}\right.\)
Xét (1):
Do \(\sqrt{3}^2+\sqrt{2}^2< \left(-\sqrt{6}\right)^2\) nên (1) vô nghiệm