giải phương trình: \(\sqrt{3}\left(cosx+2tanx\right)+sinx=\frac{3}{cosx^2}\)
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NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 9 2019
a/ ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne1\\sinx\ne-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k2\pi\\x\ne-\frac{\pi}{6}+k2\pi\\x\ne\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow cosx-sin2x=\sqrt{3}\left(1+sinx-2sin^2x\right)\)
\(\Leftrightarrow cosx-sin2x=\sqrt{3}\left(cos2x+sinx\right)\)
\(\Leftrightarrow\sqrt{3}sinx-cosx=sin2x+\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx=\frac{1}{2}sin2x+\frac{\sqrt{3}}{2}cos2x\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=sin\left(2x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow...\)
b/ ĐKXĐ: \(cosx+\sqrt{3}sinx\ne0\Leftrightarrow sin\left(x+\frac{\pi}{6}\right)\ne0\Rightarrow...\)
Đặt \(cosx+\sqrt{3}sinx=2sin\left(x+\frac{\pi}{6}\right)=a\) với \(-2\le a\le2\):
\(a=\frac{3}{a}+1\Leftrightarrow a^2-a-3=0\)
\(\Rightarrow\left[{}\begin{matrix}a=\frac{1+\sqrt{13}}{2}>2\left(l\right)\\a=\frac{1-\sqrt{13}}{2}\end{matrix}\right.\)
\(\Rightarrow2sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{13}}{2}\)
\(\Rightarrow sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{13}}{4}=sin\alpha\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=\alpha+k2\pi\\x+\frac{\pi}{6}=\pi-\alpha+k2\pi\end{matrix}\right.\) \(\Rightarrow x=...\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
\(\Leftrightarrow cos3x+\sqrt{3}sin3x=\sqrt{3}cosx+sinx\)
\(\Leftrightarrow\dfrac{1}{2}cos3x+\dfrac{\sqrt{3}}{2}sin3x=\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx\)
\(\Leftrightarrow cos\left(3x-\dfrac{\pi}{3}\right)=cos\left(x-\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\\3x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)
NL
1
KB
17 tháng 7 2021
Ta có : \(2cos^2x+2\sqrt{3}sinx.cosx+1=3\left(sinx+\sqrt{3}cosx\right)\)
\(\Leftrightarrow3cos^2x+sin^2x+2\sqrt{3}sinxcosx=3\left(sinx+\sqrt{3}cosx\right)\)
\(\Leftrightarrow\left(\sqrt{3}cosx+sinx\right)^2=3\left(\sqrt{3}cosx+sinx\right)\)
\(\Leftrightarrow\left(\sqrt{3}cosx+sinx\right)\left(\sqrt{3}cosx+sinx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3}cosx+sinx=0\\\sqrt{3}cos+sinx=3\end{matrix}\right.\)
Thấy : \(-1\le sinx;cosx\le1\Rightarrow\sqrt{3}cosx+sinx\le1+\sqrt{3}< 3\)
Do đó : \(\sqrt{3}cosx+sinx=0\) \(\Leftrightarrow\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx=0\)
\(\Leftrightarrow sin\dfrac{\pi}{3}.cosx+cos\dfrac{\pi}{3}sinx=0\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow x+\dfrac{\pi}{3}=k\pi\Leftrightarrow x=\dfrac{-\pi}{3}+k\pi\) ( k thuộc Z )
Vậy ...
NL
2
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2021
\(\Leftrightarrow\sqrt{3}sinx+cosx=\sqrt{3}\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
26 tháng 7 2021
Pt \(\Leftrightarrow sinx+\dfrac{\sqrt{3}}{3}cosx=1\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow sinx.cos\dfrac{\pi}{6}+cosx.sin\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{6}=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy...
NS
1
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 7 2021
\(\Leftrightarrow2\left(\dfrac{1}{2}cosx+\dfrac{\sqrt{3}}{2}sinx\right)+2cos\left(x-\dfrac{\pi}{3}\right)=2\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)+cos\left(x-\dfrac{\pi}{3}\right)=1\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\x-\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\pi}{3}+k2\pi\\x=k2\pi\end{matrix}\right.\)
HP
1 tháng 6 2021
1.
\(sinx-\sqrt{2}cos3x=\sqrt{3}cosx+\sqrt{2}sin3x\)
\(\Leftrightarrow sinx-\sqrt{3}cosx=\sqrt{2}cos3x+\sqrt{2}sin3x\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{\sqrt{2}}cos3x+\dfrac{1}{\sqrt{2}}sin3x\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin\left(3x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=3x+\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{3}=\pi-3x-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7\pi}{24}-k\pi\\x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm \(x=-\dfrac{7\pi}{24}-k\pi;x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\)
HP
1 tháng 6 2021
2.
\(sinx-\sqrt{3}cosx=2sin5\text{}x\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin5x\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin5x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=5x+k2\pi\\x-\dfrac{\pi}{3}=\pi-5x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2}\\x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)
MA
0
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2020
d/
ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow\frac{sin\left(3x-x\right)}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{sin2x}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{2sinx.cosx}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{sinx}{cosx}=\sqrt{3}\)
\(\Leftrightarrow tanx=\sqrt{3}\)
\(\Rightarrow x=\frac{\pi}{3}+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2020
c/
ĐKXĐ: \(sin2x\ne0\)
\(\Leftrightarrow\frac{\frac{sinx}{cosx}-sinx}{sin^3x}=\frac{1}{cosx}\)
\(\Leftrightarrow sinx-sinx.cosx=sin^3x\)
\(\Leftrightarrow1-cosx=sin^2x\)
\(\Leftrightarrow1-cosx=1-cos^2x\)
\(\Leftrightarrow cos^2x-cosx=0\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=k2\pi\end{matrix}\right.\)
ĐKXĐ: ...
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx=\dfrac{3}{2}\left(1+tan^2x\right)-\sqrt{3}tanx\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=\dfrac{3}{2}\left(tanx-\dfrac{\sqrt{3}}{3}\right)^2+1\)
\(\left\{{}\begin{matrix}sin\left(x+\dfrac{\pi}{3}\right)\le1\\\dfrac{3}{2}\left(tanx-\dfrac{\sqrt{3}}{3}\right)^2+1\ge1\end{matrix}\right.\)
Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}sin\left(x+\dfrac{\pi}{3}\right)=1\\tanx=\dfrac{\sqrt{3}}{3}\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{\pi}{6}+k2\pi\)
Hình như có nhầm lẫn từ dòng 1 xuống dòng 2 thì phải. Em bấm máy tính ra nghiệm pi/6 mà.