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NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 9 2021
Hàm là vậy phải không nhỉ? \(y=\dfrac{sin^2x-3sinx}{\left(tanx-1\right)\left(cotx+1\right)}\)
ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\tanx-1\ne0\\cotx+1\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin2x\ne0\\tanx\ne1\\cotx\ne-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{k\pi}{2}\\x\ne\dfrac{\pi}{4}+k\pi\\x\ne-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{k\pi}{2}\\x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)
\(\Leftrightarrow x\ne\dfrac{k\pi}{4}\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 10 2020
ĐKXĐ: ...
\(\Leftrightarrow1+cot^2x=cotx+3\)
\(\Leftrightarrow cot^2x-cotx-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cotx=-1\\cotx=2\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 7 2020
c/
ĐKXĐ: ...
\(\Leftrightarrow9-13cosx+4.cos^2x=0\)
\(\Leftrightarrow\left(cosx-1\right)\left(4cosx-9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{9}{4}>1\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=k2\pi\)
d/
\(\Leftrightarrow2\left(tan^2x+1\right)+1=\frac{3}{cosx}\)
\(\Leftrightarrow\frac{2}{cos^2x}-\frac{3}{cosx}+1=0\)
\(\Leftrightarrow\left(\frac{1}{cosx}-1\right)\left(\frac{2}{cosx}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\frac{1}{cosx}=1\\\frac{2}{cosx}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=k2\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 7 2020
a/
ĐKXĐ: ..
\(\Leftrightarrow1+cot^2x=cotx+3\)
\(\Leftrightarrow cot^2x-cotx-2=0\)
\(\Rightarrow\left[{}\begin{matrix}cotx=-1\\cotx=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=arccot\left(2\right)+k\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow\sqrt{3}\left(1+cot^2x\right)=3cotx+\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}cot^2x-3cotx=0\)
\(\Rightarrow\left[{}\begin{matrix}cotx=0\\cotx=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{\pi}{6}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 10 2019
1/ \(sinx=-\frac{1}{2}=sin\left(-\frac{\pi}{6}\right)\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
b/ \(cos=-\frac{\sqrt{2}}{2}=cos\left(\frac{3\pi}{4}\right)\)
\(\Rightarrow x=\pm\frac{3\pi}{4}+k2\pi\)
c/ \(tanx=\sqrt{3}=tan\left(\frac{\pi}{3}\right)\)
\(\Rightarrow x=\frac{\pi}{3}+k\pi\)
d/ \(cotx=0\Rightarrow x=\frac{\pi}{2}+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 10 2019
2/
a/ \(sin^2x+sinx-2=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(sinx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-2\left(vn\right)\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{2}+k2\pi\)
b/ \(cot^2x-2cotx-3=0\)
\(\Leftrightarrow\left(cotx+1\right)\left(cotx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cotx=-1\\cotx=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=arccot3+k\pi\end{matrix}\right.\)
3/ \(\Leftrightarrow1-cos2x+1-cos4x+1-cos6x=3\)
\(\Leftrightarrow cos2x+cos6x+cos4x=0\)
\(\Leftrightarrow2coss4x.cos2x+cos4x=0\)
\(\Leftrightarrow cos4x\left(2cos2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\2x=\frac{2\pi}{3}+k2\pi\\2x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)
Y
1
HP
17 tháng 9 2021
\(sin^2x+sin^22x=1\)
\(\Leftrightarrow2sin^2x-1+2sin^22x-2=-1\)
\(\Leftrightarrow-cos2x-2cos^22x+1=0\)
\(\Leftrightarrow\left(cos2x+1\right)\left(2cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\pi+k2\pi\\2x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pm\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2020
b/
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cosx+1-cos^2x+2cos^2x-1=\frac{1}{2}\)
\(\Leftrightarrow cos^2x+\frac{1}{2}cosx=0\)
\(\Leftrightarrow cosx\left(cosx+\frac{1}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\left(\frac{sinx}{cosx}+\frac{cosx}{sinx}\right)^2+\frac{3}{sin2x}-7=0\)
\(\Leftrightarrow\left(\frac{sin^2x+cos^2x}{sinx.cosx}\right)^2+\frac{3}{sin2x}-7=0\)
\(\Leftrightarrow\left(\frac{2}{sin2x}\right)^2+\frac{3}{sin2x}-7=0\)
Đặt \(\frac{1}{sin2x}=a\Rightarrow4a^2+3a-7=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{7}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{1}{sin2x}=1\\\frac{1}{sin2x}=-\frac{7}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{4}{7}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k2\pi\\2x=arcsin\left(-\frac{4}{7}\right)+k2\pi\\2x=\pi-arcsin\left(-\frac{4}{7}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{1}{2}arcsin\left(-\frac{4}{7}\right)+k\pi\\x=\frac{\pi}{2}-\frac{1}{2}arcsin\left(-\frac{4}{7}\right)+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2020
a/
\(\Leftrightarrow2cos2x.cosx+\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right).cos2x=0\)
\(\Leftrightarrow2cos2x.cosx+cos^22x=0\)
\(\Leftrightarrow cos2x\left(2cosx+cos2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\left(1\right)\\2cosx+cos2x=0\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x=\frac{\pi}{2}+k\pi\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
\(\left(2\right)\Leftrightarrow2cosx+2cos^2x-1=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{3}-1}{2}\\cosx=\frac{-\sqrt{3}-1}{2}< -1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm arccos\left(\frac{\sqrt{3}-1}{2}\right)+k2\pi\)
CT
1