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NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2020
c.
\(\Leftrightarrow2sin2x.cos2x+\sqrt{3}sin2x=0\)
\(\Leftrightarrow sin2x\left(2cos2x+\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=-\frac{\sqrt{3}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\2x=\frac{5\pi}{6}+k2\pi\\2x=-\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{5\pi}{12}+k\pi\\x=-\frac{5\pi}{12}+k\pi\end{matrix}\right.\)
d.
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-\sqrt{2}< -1\left(l\right)\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=-\frac{\pi}{4}+k2\pi\\2x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2020
a.
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\cosx=\frac{5}{\sqrt{3}}>1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
b.
\(\Leftrightarrow\frac{1}{2}sin4x.cos4x+\frac{1}{8}=0\)
\(\Leftrightarrow\frac{1}{4}sin8x+\frac{1}{8}=0\)
\(\Leftrightarrow sin8x=-\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}8x=-\frac{\pi}{6}+k2\pi\\8x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{48}+\frac{k\pi}{4}\\x=\frac{7\pi}{48}+\frac{k\pi}{4}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 9 2020
c.
\(\Leftrightarrow tanx=-\frac{1}{\sqrt{3}}\)
\(\Leftrightarrow x=-\frac{\pi}{6}+k\pi\)
d.
\(\Leftrightarrow\frac{1}{2}sin2x.cos2x=0\)
\(\Leftrightarrow\frac{1}{4}sin4x=0\)
\(\Leftrightarrow sin4x=0\)
\(\Leftrightarrow x=\frac{k\pi}{4}\)
e.
\(\Leftrightarrow4sin4x.cos4x=-\sqrt{2}\)
\(\Leftrightarrow2sin8x=-\sqrt{2}\)
\(\Leftrightarrow sin8x=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}8x=-\frac{\pi}{4}+k2\pi\\8x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{32}+\frac{k\pi}{4}\\x=\frac{5\pi}{32}+\frac{k\pi}{4}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 9 2020
a.
\(\Leftrightarrow sin2x=cosx\)
\(\Leftrightarrow sin2x=sin\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}-x+k2\pi\\2x=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
b.
\(\Leftrightarrow\left[{}\begin{matrix}cos\frac{x}{2}=\frac{1}{2}\\sin\frac{x}{2}=-2< -1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{2}=\frac{\pi}{3}+k2\pi\\\frac{x}{2}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k4\pi\\x=-\frac{2\pi}{3}+k4\pi\end{matrix}\right.\)
TC
GPT: \(\dfrac{\left(\sin x-\cos x\right)\left(\sin2x-3\right)-\sin2x-\cos2x+1}{2\sin x-\sqrt{2}}=0\)
1
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 1 2021
ĐKXĐ: \(sinx\ne\dfrac{\sqrt{2}}{2}\)
\(\left(sinx-cosx\right)\left(sin2x-3\right)+\left(sinx-cosx\right)^2+\left(sin^2x-cos^2x\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(sin2x-3\right)+\left(sinx-cosx\right)^2+\left(sinx-cosx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(sin2x-3+2sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\\left(sin2x-1\right)+2\left(sinx+1\right)=0\left(vô-nghiệm\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)
Kết hợp ĐKXĐ \(\Rightarrow x=-\dfrac{\pi}{4}+k2\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 7 2020
e/
ĐKXĐ: ...
\(\Leftrightarrow\frac{2sin4x.cos2x}{cos2x}-2cos4x=2\sqrt{2}\)
\(\Leftrightarrow2sin4x-2cos4x=2\sqrt{2}\)
\(\Leftrightarrow sin4x-cos4x=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(4x-\frac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sin\left(4x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow4x-\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\frac{3\pi}{16}+\frac{k\pi}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 7 2020
d/
Đặt \(sin2x-cos2x=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=t\Rightarrow\left|t\right|\le\sqrt{2}\)
\(\Rightarrow t^2-3t-4=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\2x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{3\pi}{4}+k\pi\end{matrix}\right.\)
ID
10 tháng 5 2022
\(\sin2x=\dfrac{-\sqrt{2}}{2}\Leftrightarrow\sin2x=-\dfrac{\pi}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{4}+k2\pi\\2x=\pi-\left(-\dfrac{\pi}{4}\right)+k2\pi\end{matrix}\right.=\left[{}\begin{matrix}x=-\dfrac{\pi}{8}+k\pi\\x=\dfrac{5\pi}{8}+k\pi\end{matrix}\right.\left(k\in Z\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 9 2020
3.
\(4sinx.cosx-2sinx+1-2cosx=0\)
\(\Leftrightarrow2sinx\left(2cosx-1\right)-\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
4.
\(cosx-sinx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\-4sinx.cosx=2t^2-2\end{matrix}\right.\)
Pt trở thành: \(t+2t^2-2-1=0\Leftrightarrow2t^2+t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-\frac{3}{2}< -\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}cos\left(x+\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\\x+\frac{\pi}{4}=-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 9 2020
5.
\(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x=sinx\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=x+k2\pi\\2x+\frac{\pi}{6}=\pi-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
6.
\(9sin^2x-5\left(1-sin^2x\right)-5sinx+4=0\)
\(\Leftrightarrow14sin^2x-5sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=-\frac{1}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=arcsin\left(-\frac{1}{7}\right)+k2\pi\\x=\pi-arcsin\left(-\frac{1}{7}\right)+k2\pi\end{matrix}\right.\)
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
\(\Leftrightarrow2\sin2x=\sqrt{2}\)
\(\Leftrightarrow\sin2x=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sin2x=\dfrac{\pi}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{4}+k2\pi\\2x=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+k\pi\\x=\dfrac{3\pi}{8}+k\pi\end{matrix}\right.\left(k\in Z\right)\)
it's not me OoO