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
21 tháng 9 2021
\(1-sin^23x-5sin3x+5=0\)
\(\Leftrightarrow-sin^23x-5sin3x+6=0\)
\(\Rightarrow\left[{}\begin{matrix}sin3x=1\\sin3x=-6< -1\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow3x=\dfrac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 9 2021
\(\Leftrightarrow1-sin^22x+3sin2x-3=0\)
\(\Leftrightarrow-sin^22x+3sinx-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=2>1\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow2x=\dfrac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\dfrac{\pi}{4}+k\pi\)
N
1
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 9 2021
\(\Leftrightarrow1-sin^22x-3sin2x-3=0\)
\(\Leftrightarrow sin^22x+3sin2x+2=0\)
\(\Rightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=-2< -1\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow2x=-\dfrac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\dfrac{\pi}{4}+k\pi\)
1 tháng 9 2017
a. vs m + 2
=>pttt : cos3x.cosx-sin2x+sin3xsinx+1=0
<=>\(\dfrac{1}{2}\left(cos2x+cos4x+cos2x-cos4x\right)-sin2x+1\)=0
<=>\(\dfrac{1}{2}\).2cos2x-sin2x+1=0
<=>cos2x-sin2x+1=0
<=>cos2x-sin2x-2sinxcosx+1=0
<=>cos2x+cos2x-sin2x=0
<=>2cos2x-2sinxcosx=0
<=>2cosx(cosx-sinx)=0
<=>\(\left[{}\begin{matrix}2cosx=0\\cosx-sinx=0\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{\pi}{4+k\pi}\end{matrix}\right.\)(k thuộc Z)
N
1
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 9 2021
1.
\(tan^2x-5tanx+6=0\)
\(\Rightarrow\left[{}\begin{matrix}tanx=2\\tanx=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(2\right)+k\pi\\x=arctan\left(3\right)+k\pi\end{matrix}\right.\)
2.
\(3cos^22x+4cos2x+1=0\)
\(\Rightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\pi+k2\pi\\2x=\pm arccos\left(-\dfrac{1}{3}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pm\dfrac{1}{2}arccos\left(-\dfrac{1}{3}\right)+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 6 2021
ĐKXĐ: ...
\(\Leftrightarrow\sqrt{3}\left(1+cot^2x\right)=3cotx+\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}cot^2x-3cotx=0\)
\(\Leftrightarrow\sqrt{3}cotx\left(cotx-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cotx=0\\cotx=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
HP
19 tháng 8 2021
g, \(1+2sinx=2cosx\)
\(\Leftrightarrow sinx-cosx=-\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=arcsin\left(-\dfrac{1}{2\sqrt{2}}\right)+k2\pi\\x-\dfrac{\pi}{4}=\pi-arcsin\left(-\dfrac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+arcsin\left(-\dfrac{1}{2\sqrt{2}}\right)+k2\pi\\x=\dfrac{5\pi}{4}-arcsin\left(-\dfrac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
HP
19 tháng 8 2021
h, \(4cosx-3sinx=3\)
\(\Leftrightarrow5\left(\dfrac{4}{5}cosx-\dfrac{3}{5}sinx\right)=3\)
\(\Leftrightarrow cos\left(x+arccos\dfrac{4}{5}\right)=\dfrac{3}{5}\)
\(\Leftrightarrow x+arccos\dfrac{4}{5}=\pm arccos\dfrac{3}{5}+k2\pi\)
\(\Leftrightarrow x=-arccos\dfrac{4}{5}\pm arccos\dfrac{3}{5}+k2\pi\)
HP
19 tháng 8 2021
d, \(cosx-cos2x=sin3x\)
\(\Leftrightarrow2sin\dfrac{3x}{2}.sin\dfrac{x}{2}=2sin\dfrac{3x}{2}.cos\dfrac{3x}{2}\)
\(\Leftrightarrow2sin\dfrac{3x}{2}.\left(sin\dfrac{x}{2}-cos\dfrac{3x}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\dfrac{3x}{2}=0\\sin\dfrac{x}{2}=cos\dfrac{3x}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\dfrac{3x}{2}=0\\cos\left(\dfrac{\pi}{2}-\dfrac{x}{2}\right)=cos\dfrac{3x}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x}{2}=k\pi\\\dfrac{\pi}{2}-\dfrac{x}{2}=\pm\dfrac{3x}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{4}-k\pi\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
HP
22 tháng 9 2021
5.
\(\sqrt{3}cosx-sinx=\sqrt{2}\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cosx-\dfrac{1}{2}sinx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow x+\dfrac{\pi}{6}=\pm\dfrac{\pi}{4}+k2\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k2\pi\\x=-\dfrac{5\pi}{12}+k2\pi\end{matrix}\right.\)
a, \(sin5x-cos5x+1=0\)
\(\Leftrightarrow\sqrt{2}sin\left(5x-\dfrac{\pi}{4}\right)+1=0\)
\(\Leftrightarrow sin\left(5x-\dfrac{\pi}{4}\right)=-\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\\5x-\dfrac{\pi}{4}=\pi+\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k2\pi}{5}\\x=\dfrac{3\pi}{10}+\dfrac{k2\pi}{5}\end{matrix}\right.\)
b, \(sin3x+cos3x=1\)
\(\Leftrightarrow\sqrt{2}sin\left(3x+\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(3x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\3x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\end{matrix}\right.\)