Giải pt
\(cotx-tanx=sinx+cosx\)
\(sinx+cosx+\dfrac{1}{sinx}+\dfrac{1}{cosx}=\dfrac{10}{3}\)
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NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 3 2019
Giả sử các biểu thức đã cho đều xác định
a/ \(\dfrac{1+sin^2x}{1-sin^2x}=\dfrac{1+sin^2x}{cos^2x}=\dfrac{1}{cos^2x}+\dfrac{sin^2x}{cos^2x}+1+tan^2x+tan^2x=1+2tan^2x\)
b/ \(\dfrac{sinx}{1+cosx}+\dfrac{1+cosx}{sinx}=\dfrac{sin^2x+\left(1+cosx\right)^2}{\left(1+cosx\right)sinx}=\dfrac{sin^2x+cos^2x+2cosx+1}{\left(1+cosx\right)sinx}\)
\(=\dfrac{1+2cosx+1}{\left(1+cosx\right)sinx}=\dfrac{2+2cosx}{\left(1+cosx\right)sinx}=\dfrac{2\left(1+cosx\right)}{\left(1+cosx\right)sinx}=\dfrac{2}{sinx}\)
c/ \(\dfrac{1-sinx}{cosx}=\dfrac{\left(1-sinx\right)cosx}{cos^2x}=\dfrac{\left(1-sinx\right)cosx}{1-sin^2x}\)
\(\dfrac{\left(1-sinx\right)cosx}{\left(1-sinx\right)\left(1+sinx\right)}=\dfrac{cosx}{1+sinx}\)
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 3 2019
d/ \(\left(1-cosx\right)\left(1+cot^2x\right)=\left(1-cosx\right).\dfrac{1}{sin^2x}\)
\(=\dfrac{1-cosx}{1-cos^2x}=\dfrac{1-cosx}{\left(1-cosx\right)\left(1+cosx\right)}=\dfrac{1}{1+cosx}\)
e/ \(1-\dfrac{sin^2x}{1+cotx}-\dfrac{cos^2x}{1+tanx}=1-\dfrac{sin^3x}{sinx\left(1+\dfrac{cosx}{sinx}\right)}-\dfrac{cos^3x}{cosx\left(1+\dfrac{sinx}{cosx}\right)}\)
\(=1-\left(\dfrac{sin^3x}{sinx+cosx}+\dfrac{cos^3x}{sinx+cosx}\right)=1-\left(\dfrac{sin^3x+cos^3x}{sinx+cosx}\right)\)
\(=1-\left(\dfrac{\left(sinx+cosx\right)\left(sin^2x-sinx.cosx+cos^2x\right)}{sinx+cosx}\right)\)
\(=1-\left(1-sinx.cosx\right)=sinx.cosx\)
f/ Bạn ghi đề sai à?
TA
23 tháng 12 2020
ĐK: \( \begin{cases}cotx \ne 1\\sinx \ne 0\\\end{cases} \Leftrightarrow \begin{cases}x \ne \dfrac{π}{4}+kπ\\ x \ne kπ\\\end{cases}\)
Vậy \(D=R\) \ \({\dfrac{π}{4}+kπ ; kπ | k \in Z}\)
HP
5 tháng 9 2021
1.
\(sin^3x+cos^3x=1-\dfrac{1}{2}sin2x\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)=1-sinx.cosx\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)=1-sinx.cosx\)
\(\Leftrightarrow\left(1-sinx.cosx\right)\left(sinx+cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx.cosx=1\\sinx+cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=2\left(vn\right)\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
HP
5 tháng 9 2021
2.
\(\left|cosx-sinx\right|+2sin2x=1\)
\(\Leftrightarrow\left|cosx-sinx\right|-1+2sin2x=0\)
\(\Leftrightarrow\left|cosx-sinx\right|-\left(cosx-sinx\right)^2=0\)
\(\Leftrightarrow\left|cosx-sinx\right|\left(1-\left|cosx-sinx\right|\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\\left|cosx-sinx\right|=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=k\pi\\cos^2x+sin^2x-2sinx.cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\1-sin2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\sin2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)
L
1) cosx\(^2\)+sinx=0
2) 2cos\(^2\)x-cos2x+cosx=0
3) sin\(^2\)x-3cos2x-2=0
4) tanx+\(\dfrac{2}{cotx}\)=0
2
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 9 2021
3.
\(\dfrac{1}{2}-\dfrac{1}{2}cos2x-3cos2x-2=0\)
\(\Leftrightarrow-7cos2x-3=0\)
\(\Rightarrow cos2x=-\dfrac{3}{7}\)
\(\Rightarrow2x=\pm arccos\left(-\dfrac{3}{7}\right)+k2\pi\)
\(\Rightarrow x=\pm\dfrac{1}{2}arccos\left(-\dfrac{3}{7}\right)+k\pi\)
4.
ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)
\(tanx+2tanx=0\)
\(\Rightarrow3tanx=0\)
\(\Rightarrow tanx=0\)
\(\Rightarrow x=k\pi\) (loại do ĐKXĐ)
Vậy pt đã cho vô nghiệm
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 9 2021
1.
\(\Leftrightarrow1-sin^2x+sinx=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1+\sqrt{5}}{2}>1\left(loại\right)\\sinx=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arcsin\left(\dfrac{1-\sqrt{5}}{2}\right)+k2\pi\\x=\pi-arcsin\left(\dfrac{1-\sqrt{5}}{2}\right)+k2\pi\end{matrix}\right.\) (\(k\in Z\))
2.
\(2cos^2x-\left(2cos^2x-1\right)+cosx=0\)
\(\Leftrightarrow cosx+1=0\)
\(\Leftrightarrow cosx=-1\)
\(\Leftrightarrow x=\pi+k2\pi\) (\(k\in Z\))
NM
0
PL
6 tháng 9 2023
\(P=\dfrac{sinx+cosx}{sinx-cosx}\)
\(=\dfrac{\dfrac{sinx}{cosx}+\dfrac{cosx}{cosx}}{\dfrac{sinx}{cosx}-\dfrac{cosx}{cosx}}=\dfrac{tanx+1}{tanx-1}\)
\(=\dfrac{3+1}{3-1}\)
\(=2\)
1.
ĐK: \(x\ne\dfrac{k\pi}{2}\)
\(cotx-tanx=sinx+cosx\)
\(\Leftrightarrow\dfrac{cosx}{sinx}-\dfrac{sinx}{cosx}=sinx+cosx\)
\(\Leftrightarrow\dfrac{cos^2x-sin^2x}{sinx.cosx}=sinx+cosx\)
\(\Leftrightarrow\left(\dfrac{cosx-sinx}{sinx.cosx}-1\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\left(1\right)\\cosx-sinx=sinx.cosx\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=0\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\)
\(\left(2\right)\Leftrightarrow t=\dfrac{1-t^2}{2}\left(t=cosx-sinx,\left|t\right|\le2\right)\)
\(\Leftrightarrow t^2+2t-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-1+\sqrt{2}\\t=-1-\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow cosx-sinx=-1+\sqrt{2}\)
\(\Leftrightarrow-\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=-1+\sqrt{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}-1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\\x=\dfrac{5\pi}{4}-arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm:
\(x=-\dfrac{\pi}{4}+k\pi;x=\dfrac{\pi}{4}+arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi;x=\dfrac{5\pi}{4}-arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\)