(sinx+cox)2 + 2sin2\(\dfrac{x}{2}\)=sinx( 2\(\sqrt{3}\)sinx +4-\(\sqrt{3}\))
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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.
\(cosx-\sqrt{3}sinx=1\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
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.
\(\left|cosx-sinx\right|+2sin2x=1\)
\(\Leftrightarrow\left|cosx-sinx\right|-1+2sin2x=0\)
\(\Leftrightarrow\left|cosx-sinx\right|-\left(cosx-sinx\right)^2=0\)
\(\Leftrightarrow\left|cosx-sinx\right|\left(1-\left|cosx-sinx\right|\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\\left|cosx-sinx\right|=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=k\pi\\cos^2x+sin^2x-2sinx.cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\1-sin2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\sin2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)
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 \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)
HP
25 tháng 6 2021
1. \(sin\left(\dfrac{\pi}{3}-x\right)\ne0\Leftrightarrow\dfrac{\pi}{3}-x\ne k\pi\Leftrightarrow x\ne\dfrac{\pi}{3}-k\pi\)
2. \(cos2x\ne0\Leftrightarrow2x\ne\dfrac{\pi}{2}+k\pi\Leftrightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
3. \(\sqrt{1+sinx}-\sqrt{2}\ge0\Leftrightarrow1+sinx\ge2\Leftrightarrow sinx\ge1\Leftrightarrow sinx=1\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)
4. \(\sqrt{2-2cosx}-2\ne0\Leftrightarrow2-2cosx\ne4\Leftrightarrow cosx\ne-1\Leftrightarrow x\ne\pi+k2\pi\)
5. \(1-\sqrt{1+sin3x}\ne0\Leftrightarrow sin3x\ne0\Leftrightarrow3x\ne k\pi\Leftrightarrow x\ne\dfrac{k\pi}{3}\)
\(\Leftrightarrow1+2sinx.cosx+1-cosx=2\sqrt{3}sin^2x+\left(4-\sqrt{3}\right)sinx\)
\(\Leftrightarrow cosx\left(2sinx-1\right)-\left(2\sqrt{3}sin^2x+\left(4-\sqrt{3}\right)sinx-2\right)=0\)
\(\Leftrightarrow cosx\left(2sinx-1\right)-\left(2sinx-1\right)\left(\sqrt{3}sinx+2\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(cosx+\sqrt{3}sinx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2sinx-1=0\\\dfrac{1}{2}cosx+\dfrac{\sqrt{3}}{2}sinx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\cos\left(x-\dfrac{\pi}{3}\right)=-1\end{matrix}\right.\)
\(\Leftrightarrow...\)