\(4cos^2x+2cosx-2-\sqrt{2}=0\)
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NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2020
Đặt \(cosx=t\) với \(-1\le t\le1\)
\(\Rightarrow4t^2+2t-2-\sqrt{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{-1+\sqrt{\left(2\sqrt{2}+1\right)^2}}{4}=\frac{\sqrt{2}}{2}\\t=\frac{-1-\sqrt{\left(2\sqrt{2}+1\right)^2}}{4}=\frac{-1-\sqrt{2}}{2}< -1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow cosx=\frac{\sqrt{2}}{2}\)
\(\Rightarrow x=\pm\frac{\pi}{4}+k2\pi\)
PN
1
ND
Nguyễn Đức Trí
VIP
23 tháng 7 2023
\(tanx=\dfrac{1}{cotx}=\dfrac{1}{\sqrt[]{2}}=\dfrac{\sqrt[]{2}}{2}\left(tanx.cotx=1\right)\)
\(1+tan^2x=\dfrac{1}{cos^2x}\Rightarrow cos^2x=\dfrac{1}{1+tan^2x}=\dfrac{1}{1+\dfrac{1}{2}}\)
\(\Rightarrow cos^2x=\dfrac{2}{3}\Rightarrow cosx=\sqrt[]{\dfrac{2}{3}}\)
\(tanx=\dfrac{sinx}{cosx}\Rightarrow sinx=tanx.cosx=\dfrac{1}{\sqrt[]{2}}.\dfrac{\sqrt[]{2}}{\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{3}\)
\(P=\dfrac{3sinx-2cosx}{12sin^3x+4cos^3x}=\dfrac{3.\dfrac{\sqrt[]{3}}{3}-2.\dfrac{\sqrt[]{2}}{\sqrt[]{3}}}{12.\left(\dfrac{\sqrt[]{3}}{3}\right)^3+4.\left(\sqrt[]{\dfrac{2}{3}}\right)^3}\)
\(=\dfrac{\sqrt[]{3}-\dfrac{2\sqrt[]{6}}{3}}{12.\left(\dfrac{\sqrt[]{3}}{3}\right)^3+4.\left(\sqrt[]{\dfrac{2}{3}}\right)^3}\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 8 2020
e/
\(\Leftrightarrow\left(sin^2x+4sinx.cosx+3cos^2x\right)-\left(sinx+3cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sinx+3cosx\right)-\left(sinx+3cosx\right)=0\)
\(\Leftrightarrow\left(sinx+3cosx\right)\left(sinx+cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+3cosx=0\\sinx+cosx-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-3cosx\\\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-3\\sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arctan\left(-3\right)+k\pi\\x=k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 8 2020
d/
\(\Leftrightarrow2sinx+2sinx.cos2x-\left(1-sin2x\right)-2cosx=0\)
\(\Leftrightarrow2\left(sinx-cosx\right)+2sinx\left(cos^2x-sin^2x\right)-\left(sinx-cosx\right)^2=0\)
\(\Leftrightarrow2\left(sinx-cosx\right)-2sinx\left(sinx-cosx\right)\left(sinx+cosx\right)-\left(sinx-cosx\right)^2=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(2-2sin^2x-2sinx.cosx-sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left[2cos^2x-2sinx.cosx-sinx+cosx\right]=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left[2cosx\left(cosx-sinx\right)+cosx-sinx\right]=0\)
\(\Leftrightarrow-\left(sinx-cosx\right)^2\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\2cosx+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
NH
\(\dfrac{2sin^3x+2\sqrt{3}sin^2x.cosx-2sin^2x+cos\left(2x+\dfrac{\pi}{3}\right)}{2cosx-\sqrt{3}}=0\)
0
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 7 2020
d/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+4=4\left(\sqrt{3}sinx+cosx\right)\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+\frac{5}{2}=4\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow2sin^2\left(x+\frac{\pi}{6}\right)+4sin\left(x+\frac{\pi}{6}\right)-\frac{7}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{-2+\sqrt{11}}{2}\\sin\left(x+\frac{\pi}{6}\right)=\frac{-2-\sqrt{11}}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+arcsin\left(\frac{-2+\sqrt{11}}{2}\right)+k2\pi\\x=\frac{5\pi}{6}-arcsin\left(\frac{-2+\sqrt{11}}{2}\right)+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 7 2020
c/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=...\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 8 2020
d/ ĐKXĐ: ...
\(\Leftrightarrow\frac{\left(cosx-sinx\right)\left(cos^2x+sin^2x+sinx.cosx\right)}{2cosx+3sinx}=cos^2x-sin^2x\)
\(\Leftrightarrow\frac{\left(cosx-sinx\right)\left(1+sinx.cosx\right)}{2cosx+3sinx}=\left(cosx-sinx\right)\left(cosx+sinx\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx-sinx=0\Leftrightarrow x=\frac{\pi}{4}+k\pi\\\frac{1+sinx.cosx}{2cosx+3sinx}=sinx+cosx\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow1+sinx.cosx=\left(sinx+cosx\right)\left(2cosx+3sinx\right)\)
\(\Leftrightarrow1+sinx.cosx=2sin^2x+3cos^2x+5sinx.cosx\)
\(\Leftrightarrow2sin^2x+3cos^2x+4sinx.cosx-1=0\)
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)
\(2tan^2x+3+4tanx-1-tan^2x=0\)
\(\Leftrightarrow tan^2x+4tanx+2=0\)
\(\Leftrightarrow tanx=-2\pm\sqrt{2}\)
\(\Rightarrow x=arctan\left(-2\pm\sqrt{2}\right)+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 8 2020
c/
\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx+4cosx\right)=4\left(sinx-cosx\right)\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx+4cosx-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\left(1\right)\\sinx+4cosx-4=0\left(2\right)\end{matrix}\right.\)
Xét (1) \(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=0\Leftrightarrow x=\frac{\pi}{4}+k\pi\)
Xét (2) \(\Leftrightarrow\frac{1}{\sqrt{17}}sinx+\frac{4}{\sqrt{17}}cosx=\frac{4}{\sqrt{17}}\)
Đặt \(\frac{4}{\sqrt{17}}=cosa\) với \(a\in\left(0;\pi\right)\)
\(\Rightarrow cosx.cosa+sinx.sina=cosa\)
\(\Leftrightarrow cos\left(x-a\right)=cosa\)
\(\Leftrightarrow\left[{}\begin{matrix}x-a=a+k2\pi\\x-a=-a+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2a+k2\pi\\x=k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 6 2021
1.
\(\Leftrightarrow2sinx.cosx+2cosx=0\)
\(\Leftrightarrow2cosx\left(sinx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=-1\end{matrix}\right.\)
\(\Leftrightarrow cosx=0\) (do \(cosx=0\Leftrightarrow sinx=\pm1\) bao hàm luôn cả pt \(sinx=-1\))
\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)
2.
\(\Leftrightarrow\left[{}\begin{matrix}2x-10^0=60^0+k360^0\\2x-10^0=120^0+n360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=35^0+k180^0\\x=65^0+n180^0\end{matrix}\right.\)
Do \(-120^0< x< 90^0\Rightarrow\left\{{}\begin{matrix}-120^0< 35^0+k180^0< 90^0\\-120^0< 65^0+n180^0< 90^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}k=0\\n=\left\{-1;0\right\}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=35^0\\x=-115^0\\x=65^0\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 6 2021
3. Làm tương tự câu 2
4.
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos\left(10x+\dfrac{4\pi}{5}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}cos\left(\dfrac{x}{2}-2\pi\right)\right)=0\)
\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}-2\pi\right)=0\)
\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)=-cos\left(\dfrac{x}{2}\right)=cos\left(\pi-\dfrac{x}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}10x+\dfrac{4\pi}{5}=\pi-\dfrac{x}{2}+k2\pi\\10x+\dfrac{4\pi}{5}=\dfrac{x}{2}-\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
LK
1
17 tháng 6 2016
điều kiện : cosx\(\ne\)\(\frac{1}{\sqrt{2}}\)=> x\(\ne\)\(\pm\)\(\frac{\pi}{4}\)+2k\(\pi\), k\(\in\)Z
pt<=> tử số =0
<=>cos2x-sin(3x-\(\frac{\pi}{4}\)+x+\(\frac{3\pi}{4}\))-sin(3x-\(\frac{\pi}{4}\)-x-\(\frac{3\pi}{4}\))-2=0
<=> cos2x-sin(x+\(\frac{\pi}{2}\))-sin(2x-\(\pi\))-2=0
<=> cos2x-cosx+sin2x-2sin2x-2cos2x=0
<=>-cos2x-coxs+2sinx.cosx-2sin2x=0
đến đây bạn nhóm lại ra nghiệm rồi kiểm tra đk là xong
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 7 2021
1a.
Đặt \(5x+6=u\)
\(cos2u+4\sqrt{2}sinu-4=0\)
\(\Leftrightarrow1-2sin^2u+4\sqrt{2}sinu-4=0\)
\(\Leftrightarrow2sin^2u-4\sqrt{2}sinu+3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinu=\dfrac{3\sqrt{2}}{2}>1\left(loại\right)\\sinu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow sin\left(5x+6\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+6=\dfrac{\pi}{4}+k2\pi\\5x+6=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}+\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=-\dfrac{6}{5}+\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 7 2021
1b.
Đặt \(2x+1=u\)
\(cos2u+3sinu=2\)
\(\Leftrightarrow1-2sin^2u+3sinu=2\)
\(\Leftrightarrow2sin^2u-3sinu+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinu=1\\sinu=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(2x+1\right)=1\\sin\left(2x+1\right)=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{\pi}{2}+k2\pi\\2x+1=\dfrac{\pi}{6}+k2\pi\\2x+1=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}+\dfrac{\pi}{4}+k\pi\\x=-\dfrac{1}{2}+\dfrac{\pi}{12}+k\pi\\x=-\dfrac{1}{2}+\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)
Câu này và câu bạn vừa đăng về bản chất là một
Bạn chia 2 vế của pt cho 2 nó sẽ thành giống hệt bài trước ko khác 1 chữ luôn