(sqrt(sinx+4)-sqr(sinx))(1+sqrt(sinx^sinx+4sinx))=4
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DS
1
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 9 2021
\(\Leftrightarrow2cos2x.cos\left(\dfrac{\pi}{4}\right)+4sinx=2+\sqrt{2}-\sqrt{2}sinx\)
\(\Leftrightarrow\sqrt{2}cos2x+\left(4+\sqrt{2}\right)sinx=2+\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}\left(1-2sin^2x\right)+\left(4+\sqrt{2}\right)sinx=2+\sqrt{2}\)
\(\Leftrightarrow-2\sqrt{2}sin^2x+\left(4+\sqrt{2}\right)sinx-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\sqrt{2}>1\left(loại\right)\\sinx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
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 \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)
ND
1
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 9 2021
\(\Leftrightarrow1+2sinx.cosx+1-cosx=2\sqrt{3}sin^2x+\left(4-\sqrt{3}\right)sinx\)
\(\Leftrightarrow cosx\left(2sinx-1\right)-\left(2\sqrt{3}sin^2x+\left(4-\sqrt{3}\right)sinx-2\right)=0\)
\(\Leftrightarrow cosx\left(2sinx-1\right)-\left(2sinx-1\right)\left(\sqrt{3}sinx+2\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(cosx+\sqrt{3}sinx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2sinx-1=0\\\dfrac{1}{2}cosx+\dfrac{\sqrt{3}}{2}sinx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\cos\left(x-\dfrac{\pi}{3}\right)=-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
NT
1
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 4 2021
\(=\dfrac{1+sinx+1-sinx}{\sqrt{\left(1-sinx\right)\left(1+sinx\right)}}=\dfrac{2}{\sqrt{1-sin^2x}}=\dfrac{2}{\sqrt{cos^2x}}=\dfrac{2}{\left|cosx\right|}\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
2.1
a.
\(\Leftrightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
b.
\(cosx-\sqrt{3}sinx=1\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
LT
1
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 4 2019
\(\sqrt{\frac{1+sinx}{1-sinx}}+\sqrt{\frac{1-sinx}{1+sinx}}=\sqrt{\frac{sin^2\frac{x}{2}+cos^2\frac{x}{2}+2sin\frac{x}{2}.cos\frac{x}{2}}{sin^2\frac{x}{2}+cos^2\frac{x}{2}-2sin\frac{x}{2}.cos\frac{x}{2}}}+\sqrt{\frac{sin^2\frac{x}{2}+cos^2\frac{x}{2}-2sin\frac{x}{2}.cos\frac{x}{2}}{sin^2\frac{x}{2}+cos^2\frac{x}{2}+2sin\frac{x}{2}.cos\frac{x}{2}}}\)
\(=\sqrt{\frac{\left(sin\frac{x}{2}+cos\frac{x}{2}\right)^2}{\left(sin\frac{x}{2}-cos\frac{x}{2}\right)^2}}+\sqrt{\frac{\left(sin\frac{x}{2}-cos\frac{x}{2}\right)^2}{\left(sin\frac{x}{2}+cos\frac{x}{2}\right)^2}}=\frac{\left|sin\frac{x}{2}+cos\frac{x}{2}\right|}{\left|sin\frac{x}{2}-cos\frac{x}{2}\right|}+\frac{\left|sin\frac{x}{2}-cos\frac{x}{2}\right|}{\left|sin\frac{x}{2}+cos\frac{x}{2}\right|}\)
\(=\frac{\left(sin\frac{x}{2}+cos\frac{x}{2}\right)^2+\left(sin\frac{x}{2}-cos\frac{x}{2}\right)^2}{\left|sin^2\frac{x}{2}-cos^2\frac{x}{2}\right|}=\frac{2}{\left|cosx\right|}\)
20 tháng 9 2020
ĐK: ...
\(VT=\left[\frac{\left(1+sinx\right)-\left(1-sinx\right)}{\sqrt{1-sin^2x}}\right]^2=\left(\frac{2sinx}{cosx}\right)^2=4tan^2x=VP\left(đpcm\right)\)