K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
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.
\(\left|cosx-sinx\right|+2sin2x=1\)
\(\Leftrightarrow\left|cosx-sinx\right|-1+2sin2x=0\)
\(\Leftrightarrow\left|cosx-sinx\right|-\left(cosx-sinx\right)^2=0\)
\(\Leftrightarrow\left|cosx-sinx\right|\left(1-\left|cosx-sinx\right|\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\\left|cosx-sinx\right|=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=k\pi\\cos^2x+sin^2x-2sinx.cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\1-sin2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\sin2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)
NM
1
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 4 2021
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: \(\left\{{}\begin{matrix}sin\left(x+\dfrac{\pi}{3}\right)=1\\tanx=\dfrac{\sqrt{3}}{3}\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{\pi}{6}+k2\pi\)
3 tháng 4 2021
Hình như có nhầm lẫn từ dòng 1 xuống dòng 2 thì phải. Em bấm máy tính ra nghiệm pi/6 mà.
DH
1
HP
17 tháng 10 2021
ĐK: \(x\ne\dfrac{\pi}{6}+k2\pi;x\ne\dfrac{5\pi}{6}+k2\pi\)
\(\dfrac{cosx-\sqrt{3}sinx}{sinx-\dfrac{1}{2}}=0\)
\(\Leftrightarrow cosx-\sqrt{3}sinx=0\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=0\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow x+\dfrac{\pi}{3}=\dfrac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)
Đối chiếu điều kiện ta được \(x=-\dfrac{5\pi}{6}+k2\pi\).
NM
1
14 tháng 8 2023
=>(cosx+sinx)-2*sinx*cosx*(sinx+cosx)=0
=>\(\left(sinx+cosx\right)\left(2\cdot sinx\cdot cosx-1\right)=0\)
=>\(\sqrt{2}\cdot sin\left(x+\dfrac{pi}{4}\right)\cdot\left(sin2x-1\right)=0\)
=>\(\left[{}\begin{matrix}sin\left(x+\dfrac{pi}{4}\right)=0\\sin2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{pi}{4}=kpi\\sin2x=1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=kpi-\dfrac{pi}{4}\\2x=\dfrac{pi}{2}+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=kpi-\dfrac{pi}{4}\\x=\dfrac{pi}{4}+kpi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
\(\Leftrightarrow cos3x+\sqrt{3}sin3x=\sqrt{3}cosx+sinx\)
\(\Leftrightarrow\dfrac{1}{2}cos3x+\dfrac{\sqrt{3}}{2}sin3x=\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx\)
\(\Leftrightarrow cos\left(3x-\dfrac{\pi}{3}\right)=cos\left(x-\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\\3x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=cos3x\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=cos3x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=x+\dfrac{\pi}{3}+k2\pi\\3x=-x-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{12}+\dfrac{k\pi}{2}\end{matrix}\right.\)
HT
0
HP
24 tháng 12 2021
\(\sqrt{3}cos2x-sin2x=\sqrt{3}sinx+cosx\)
\(\Leftrightarrow\sqrt{3}cos2x-\sqrt{3}sinx-sin2x-cosx=0\)
\(\Leftrightarrow\sqrt{3}\left(1-2sin^2x-sinx\right)-2sinx.cosx-cosx=0\)
\(\Leftrightarrow-\sqrt{3}\left(sinx+1\right)\left(2sinx-1\right)-cosx\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left[\sqrt{3}\left(sinx+1\right)+cosx\right]=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(\sqrt{3}sinx+cosx+\sqrt{3}\right)=0\)
\(\Leftrightarrow sinx=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cosx-\dfrac{1}{2}sinx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{6}\right)=cos\left(\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{6}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{6}=-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k2\pi\\x=-\dfrac{5\pi}{12}+k2\pi\end{matrix}\right.\)