Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(sin\left(x+\frac{\pi}{6}\right)=-\frac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=-\frac{\pi}{3}+k2\pi\\x+\frac{\pi}{6}=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{2}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
Do \(x\in\left[0;2\pi\right]\Rightarrow\left[{}\begin{matrix}0\le-\frac{\pi}{2}+k2\pi\le2\pi\\0\le\frac{7\pi}{6}+k2\pi\le2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}k=1\\k=0\end{matrix}\right.\) \(\Rightarrow x=\left\{\frac{3\pi}{2};\frac{7\pi}{6}\right\}\)
\(\sqrt{3}cosx+2sin^2\left(\dfrac{x}{2}-\pi\right)=1\)
\(\Leftrightarrow\sqrt{3}cosx+2sin^2\dfrac{x}{2}=1\)
\(\Leftrightarrow\sqrt{3}cosx-cosx=0\Leftrightarrow cosx=0\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\) ( k thuộc Z )
Vậy ...
22.
Nhận thấy \(cosx=0\) không phải nghiệm, chia 2 vế cho \(cos^2x\)
\(3tan^2x+2tanx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(\dfrac{1}{3}\right)+k\pi\end{matrix}\right.\)
Nghiệm dương nhỏ nhất của pt là: \(x=arctan\left(\dfrac{1}{3}\right)\)
\(\Leftrightarrow2cos^2\left(x+\frac{\pi}{3}\right)-1+3cos\left(x+\frac{\pi}{3}\right)+2=0\)
\(\Leftrightarrow2cos^2\left(x+\frac{\pi}{3}\right)+3cos\left(x+\frac{\pi}{3}\right)+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x+\frac{\pi}{3}\right)=-1\\cos\left(x+\frac{\pi}{3}\right)=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\pi+k2\pi\\x+\frac{\pi}{3}=\frac{2\pi}{3}+k2\pi\\x+\frac{\pi}{3}=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
\(cos\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\\x+\frac{\pi}{4}=-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=-\pi+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=\left\{\frac{\pi}{2};\pi\right\}\)
c/ ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow tan^3x+1+tan^2x+4\sqrt{3}\left(1+tanx\right)=8+7tanx\)
\(\Leftrightarrow tan^2x\left(1+tanx\right)+\left(4\sqrt{3}-7\right)\left(1+tanx\right)=0\)
\(\Leftrightarrow\left(tan^2x-7+4\sqrt{3}\right)\left(1+tanx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tan^2x=7-4\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=2-\sqrt{3}\\tanx=-2+\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tanx=tan\left(-\frac{\pi}{4}\right)\\tanx=tan\left(\frac{\pi}{12}\right)\\tanx=tan\left(-\frac{\pi}{12}\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{\pi}{12}+k\pi\\x=-\frac{\pi}{12}+k\pi\end{matrix}\right.\)
Bạn tự tìm x thuộc khoảng đã cho
b/
ĐKXĐ: \(cos2x\ne0\)
\(\Leftrightarrow tan^22x+1+tan^22x=7\)
\(\Leftrightarrow tan^22x=3\)
\(\Rightarrow\left[{}\begin{matrix}tan2x=\sqrt{3}\\tan2x=-\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tan2x=tan60^0\\tan2x=tan\left(-60^0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=60^0+k180^0\\2x=-60^0+k180^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=30^0+k180^0\\x=-30^0+k180^0\end{matrix}\right.\)
Bạn tự tìm nghiệm thuộc khoảng đã cho nhé
\(\sqrt{2}cos\left(x+\dfrac{\pi}{3}\right)=1\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow x+\dfrac{\pi}{3}=\pm\dfrac{\pi}{4}+k2\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k2\pi\\x=-\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\)
\(0\le-\dfrac{\pi}{12}+k2\pi\le2\pi\Leftrightarrow...\)
\(0\le-\dfrac{7\pi}{12}+k2\pi\le2\pi\Leftrightarrow...\)
Mình vội nên suy nghĩ có 5 phút nếu sai sót gì mong bạn thông cảm
\(sinx+cos\left(2x+\dfrac{\Omega}{3}\right)=0\)
=>\(cos\left(2x+\dfrac{\Omega}{3}\right)=-sinx=sin\left(-x\right)\)
=>\(cos\left(2x+\dfrac{\Omega}{3}\right)=cos\left(\dfrac{\Omega}{2}+x\right)\)
=>\(\left[{}\begin{matrix}2x+\dfrac{\Omega}{3}=x+\dfrac{\Omega}{2}+k2\Omega\\2x+\dfrac{\Omega}{3}=-x-\dfrac{\Omega}{2}+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{\Omega}{6}+k2\Omega\\3x=-\dfrac{5}{6}\Omega+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{5}{6}\Omega+k2\Omega\\x=-\dfrac{5}{18}\Omega+\dfrac{k2\Omega}{3}\end{matrix}\right.\)
TH1: \(x=\dfrac{5}{6}\Omega+k2\Omega\)
\(0< =x< =2\Omega\)
=>\(0< =\dfrac{5}{6}\Omega+k2\Omega< =2\Omega\)
=>\(-\dfrac{5}{6}\Omega< =k2\Omega< =\dfrac{7}{6}\Omega\)
=>\(-\dfrac{5}{6}< =2k< =\dfrac{7}{6}\)
=>-5/12<=k<=7/12
mà k nguyên
nên k=0
TH2: \(x=-\dfrac{5}{18}\Omega+\dfrac{k2\Omega}{3}\)
\(0< =x< =2\Omega\)
=>\(0< =-\dfrac{5}{18}\Omega+\dfrac{k2\Omega}{3}< =2\Omega\)
=>\(\dfrac{5}{18}\Omega< =\dfrac{k2\Omega}{3}< =\dfrac{41}{18}\Omega\)
=>\(\dfrac{5}{18}< =\dfrac{2k}{3}< =\dfrac{41}{18}\)
=>\(\dfrac{5}{6}< =2k< =\dfrac{41}{6}\)
=>\(\dfrac{5}{12}< =k< =\dfrac{41}{12}\)
mà k nguyên
nên \(k\in\left\{1;2;3\right\}\)
=>Có 4 nghiệm thỏa mãn
\(\Leftrightarrow2x+\frac{\pi}{3}=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{12}+\frac{k\pi}{2}\)
Do \(x\in\left[0;2\pi\right]\Rightarrow0\le\frac{\pi}{12}+\frac{k\pi}{2}\le2\pi\)
\(\Rightarrow-\frac{1}{6}\le k\le\frac{23}{6}\Rightarrow k=\left\{0;1;2;3\right\}\)
\(\Rightarrow x=\left\{\frac{\pi}{12};\frac{7\pi}{12};\frac{13\pi}{12};\frac{19\pi}{12}\right\}\)