Giải pt
\(2sin^22x+sin7x-1=sinx\)
K
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NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 8 2020
c/
\(\Leftrightarrow2cos4x.sin3x=cos4x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\2sin3x=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\sin3x=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\3x=\frac{\pi}{6}+k2\pi\\3x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
d/
\(\Leftrightarrow6sinx+3cosx+3=sinx-2cosx+3\)
\(\Leftrightarrow sinx+cosx=0\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=0\Leftrightarrow x=-\frac{\pi}{4}+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 8 2020
a/
\(\Leftrightarrow\frac{\sqrt{3}}{2}cosx-\frac{1}{2}sinx=sin4x\)
\(\Leftrightarrow sin\left(\frac{\pi}{3}-x\right)=sin4x\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{3}-x+k2\pi\\4x=\frac{2\pi}{3}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{15}+\frac{k2\pi}{5}\\x=\frac{2\pi}{9}+\frac{k2\pi}{3}\end{matrix}\right.\)
b/
\(\Leftrightarrow2sinx.cosx+4sinx.cos^2x-2sinx=0\)
\(\Leftrightarrow2sinx\left(cosx+2cos^2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\2cos^2x+cosx-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=-1\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2020
d/
ĐKXĐ: ...
\(\Leftrightarrow tanx-1+cos2x=0\)
\(\Leftrightarrow\frac{sinx}{cosx}-1-\left(sin^2x-cos^2x\right)=0\)
\(\Leftrightarrow\frac{sinx-cosx}{cosx}-\left(sinx-cosx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(\frac{1}{cosx}-sinx-cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\left(1\right)\\\frac{1}{cosx}-sinx-cosx=0\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=0\)
\(\Rightarrow x-\frac{\pi}{4}=k\pi\Rightarrow x=\frac{\pi}{4}+k\pi\)
\(\left(2\right)\Leftrightarrow1-sinx.cosx-cos^2x=0\)
\(\Leftrightarrow sin^2x-sinx.cosx=0\)
\(\Leftrightarrow sinx\left(sinx-cosx\right)=0\)
\(\Leftrightarrow sinx=0\Rightarrow x=k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2020
c/
\(\Leftrightarrow sinx.cos2x-sinx+1-cos2x=0\)
\(\Leftrightarrow sinx\left(cos2x-1\right)-\left(cos2x-1\right)=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\cos2x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\2x=k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=k\pi\end{matrix}\right.\)
K
1
24 tháng 6 2021
a) pt <=> - cos2x. tan22x + 3.cos2x=0
<=> \(\dfrac{sin^22x}{-cos2x}\)+ 3cos2x =0
<=> sin22x - 3cos22x = 0
<=> 1 - 4 cos22x = 0
<=> 1 - 4.\(\dfrac{1+cos4x}{2}\)= 0
<=> cos4x = \(\dfrac{-1}{2}\)
HP
1 tháng 6 2021
1.
\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)
\(\Leftrightarrow\sqrt{3}sinx+cosx+sinx+2cosx=3\)
\(\Leftrightarrow\left(\sqrt{3}+1\right)sinx+3cosx=3\)
\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)
Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)
\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)
\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm:
\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)
HP
1 tháng 6 2021
2.
\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)
\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)
\(\Leftrightarrow\left(2cos^2x-1\right)sinx+cos2x.cosx+2cos2x=0\)
\(\Leftrightarrow cos2x.sinx+cos2x.cosx+2cos2x=0\)
\(\Leftrightarrow cos2x.\left(sinx+cosx+2\right)=0\)
\(\Leftrightarrow cos2x=0\)
\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 8 2020
\(\Leftrightarrow2cosx-sinx-4sin^2x.cosx+2sin^3x=sin^3x+cos^3x\)
\(\Leftrightarrow sin^3x-cos^3x-4sin^2x.cosx+2cosx-sinx=0\)
- Với \(\left\{{}\begin{matrix}cosx=0\\sinx=1\end{matrix}\right.\) \(\Leftrightarrow x=\frac{\pi}{2}+k2\pi\) là nghiệm của pt
- Với \(cosx\ne0\) chia 2 vế cho \(cos^3x\)
\(tan^3x-1-4tan^2x+2\left(1+tan^2x\right)-tanx\left(1+tan^2x\right)=0\)
\(\Leftrightarrow-2tan^2x-tanx+3=0\)
\(\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=-\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=arctan\left(-\frac{3}{2}\right)+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2021
\(\Leftrightarrow2\left(1-cos^22x\right)=2+\left(1-2sin^2x\right)\)
\(\Leftrightarrow2-2cos^22x=2+cos2x\)
\(\Leftrightarrow2cos^22x+cos2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\2x=\pm\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=\pm\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow1-cos4x+sin7x-1=sinx\)
\(\Leftrightarrow sin7x-sinx-cos4x=0\)
\(\Leftrightarrow2.cos4x.sin3x-cos4x=0\)
\(\Leftrightarrow cos4x\left(2.sin3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\sin3x=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{2}+k\pi\\3x=\dfrac{\pi}{6}+k2\pi\\3x=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)(\(k\in Z\)) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\) (\(k\in Z\))
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