giải pt: cot ( 2x - 3 ) = 1/4
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
8 tháng 9 2019
*công thức : tan x = cot (\(\frac{\pi}{2}\)- x) áp dụng vào làm bạn nhé
HT
31 tháng 8 2019
a/ \(\tan^2x-\cot^2\left(x-\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow\frac{1}{\cos^2x}-1-\frac{1}{\sin^2\left(x-\frac{\pi}{4}\right)}+1=0\)
\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\left(\sin x.\cos\frac{\pi}{4}-\cos x.\sin\frac{\pi}{4}\right)^2}=0\)
\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\left(\frac{\sqrt{2}}{2}\sin x-\frac{\sqrt{2}}{2}\cos x\right)^2}=0\)
\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\frac{1}{2}\sin^2x-\sin x.\cos x+\frac{1}{2}\cos^2x}=0\)
\(\Leftrightarrow\frac{1}{2}\sin^2x-\sin x.\cos x+\frac{1}{2}\cos^2x-\cos^2x=0\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}\cos^2x-\sin x.\cos x-\frac{1}{2}\cos^2x=0\)
\(\Leftrightarrow\cos^2x+\sin x.\cos x-\frac{1}{2}=0\)
Đến đây là dễ r nha bn :3
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 10 2020
b.
\(\Leftrightarrow\frac{2\pi}{3}\left(sinx-1\right)=k2\pi\)
\(\Leftrightarrow sinx-1=3k\)
\(\Leftrightarrow sinx=3k+1\)
Do \(-1\le sinx\le1\)
\(\Rightarrow-1\le3k+1\le1\Rightarrow-\frac{2}{3}\le k\le0\)
\(\Rightarrow k=0\)
\(\Rightarrow sinx=1\)
\(\Rightarrow x=\frac{\pi}{2}+k2\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 10 2020
c.
ĐKXĐ: ...
\(\Leftrightarrow\frac{\pi}{4}\left(cosx-1\right)=-\frac{\pi}{4}+k\pi\)
\(\Leftrightarrow cosx-1=4k-1\)
\(\Leftrightarrow cosx=4k\)
Mà \(-1\le cosx\le1\Rightarrow-1\le4k\le1\)
\(\Rightarrow-\frac{1}{4}\le k\le\frac{1}{4}\Rightarrow k=0\)
\(\Rightarrow cosx=0\)
\(\Rightarrow x=\frac{\pi}{2}+k\pi\)
NT
1
HT
24 tháng 8 2020
\(DKXD:\left\{{}\begin{matrix}\cos\left(2x+\frac{\pi}{8}\right)\ne0\\\sin\left(x-\frac{3\pi}{4}\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+\frac{\pi}{8}\ne\frac{\pi}{2}+k\pi\\x-\frac{3\pi}{4}\ne k\pi\end{matrix}\right.\)
\(pt\Leftrightarrow\tan\left(2x+\frac{\pi}{8}\right)=-\cot\left(x-\frac{3\pi}{4}\right)=\tan\left(x-\frac{3\pi}{4}+\frac{\pi}{2}\right)\)
\(\Leftrightarrow2x+\frac{\pi}{8}=x-\frac{3\pi}{4}+\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=-\frac{3}{8}\pi+k\pi\)
6 tháng 12 2022
a: \(\Leftrightarrow cos2x=\dfrac{1}{\sqrt{2}}\)
=>2x=pi/4+k2pi hoặc 2x=-pi/4+k2pi
=>x=pi/8+kpi hoặc x=-pi/8+kpi
b: \(\Leftrightarrow sinx=sin\left(\dfrac{pi}{2}-3x\right)\)
=>x=pi/2-3x+k2pi hoặ x=pi/2+3x+k2pi
=>4x=pi/2+k2pi hoặc -2x=pi/2+k2pi
=>x=pi/8+kpi/2 hoặc x=-pi/4-kpi
d: \(\Leftrightarrow cos\left(x+\dfrac{pi}{3}\right)=-sin\left(3x+\dfrac{pi}{4}\right)\)
\(\Leftrightarrow cos\left(x+\dfrac{pi}{3}\right)=sin\left(-3x-\dfrac{pi}{4}\right)\)
\(\Leftrightarrow cos\left(x+\dfrac{pi}{3}\right)=cos\left(3x+\dfrac{3}{4}pi\right)\)
=>3x+3/4pi=x+pi/3+k2pi hoặc 3x+3/4pi=-x-pi/3+k2pi
=>2x=-5/12pi+k2pi hoặc 4x=-13/12pi+k2pi
=>x=-5/24pi+kpi hoặc x=-13/48pi+kpi/2
e: \(\Leftrightarrow sinx-\sqrt{3}\cdot cosx=0\)
\(\Leftrightarrow sin\left(x-\dfrac{pi}{3}\right)=0\)
=>x-pi/3=kpi
=>x=kpi+pi/3
5 tháng 8 2017
\(tan\cdot\left(x+\dfrac{\pi}{4}\right)+cot\cdot\left(2x-\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=-cot\cdot\left(2x-\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=cot\cdot\left(-2x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=tan\cdot\left(\dfrac{\pi}{2}+2x-\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=tan\cdot\left(\dfrac{\pi}{6}+2x\right)\)
\(\Leftrightarrow x+\dfrac{\pi}{4}=\dfrac{\pi}{6}+2x+k\pi\)
\(\Leftrightarrow-x=\dfrac{-\pi}{12}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{12}-k\pi\left(k\in Z\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 9 2021
1.
\(sin\left(4x-10^0\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(4x-10^0\right)=sin45^0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-10^0=45^0+k360^0\\4x-10^0=135^0+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=55^0+k360^0\\4x=145^0+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=13,75^0+k90^0\\x=36,25^0+k90^0\end{matrix}\right.\) (\(k\in Z\))
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 9 2021
2.
Đề không đúng
3.
ĐKXĐ: \(\left\{{}\begin{matrix}cos2x\ne0\\cosx\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(tan2x=tanx\)
\(\Rightarrow2x=x+k\pi\)
\(\Rightarrow x=k\pi\)
4.
\(cot\left(x+\dfrac{\pi}{5}\right)=-1\)
\(\Leftrightarrow x+\dfrac{\pi}{5}=-\dfrac{\pi}{4}+k\pi\)
\(\Leftrightarrow x=-\dfrac{9\pi}{20}+k\pi\) (\(k\in Z\))
\(\Leftrightarrow2x-3=arccot\left(\dfrac{1}{4}\right)+k\pi\)
\(\Rightarrow x=\dfrac{3}{2}+\dfrac{1}{2}arccot\left(\dfrac{1}{4}\right)+\dfrac{k\pi}{2}\)