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NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 10 2020
1.
\(\Leftrightarrow3x=k\pi\Leftrightarrow x=\frac{k\pi}{3}\)
2.
\(\Leftrightarrow cos5x=0\Leftrightarrow5x=\frac{\pi}{2}+k\pi\Leftrightarrow x=\frac{\pi}{10}+\frac{k\pi}{5}\)
4.
\(cos3x+cosx+cos2x=0\)
\(\Leftrightarrow2cos2x.cosx+cos2x=0\)
\(\Leftrightarrow cos2x\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cosx=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 10 2020
5.
\(sin6x+sin2x+sin4x=0\)
\(\Leftrightarrow2sin4x.cos2x+sin4x=0\)
\(\Leftrightarrow sin4x\left(2cos2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\cos2x=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{4}\\x=\pm\frac{\pi}{3}+k\pi\end{matrix}\right.\)
6. ĐKXĐ; ...
\(\Leftrightarrow tanx+tan2x=1-tanx.tan2x\)
\(\Leftrightarrow\frac{tanx+tan2x}{1-tanx.tan2x}=1\)
\(\Leftrightarrow tan3x=1\)
\(\Leftrightarrow x=\frac{\pi}{12}+\frac{k\pi}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2020
c.
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos\left(8x+\frac{2\pi}{3}\right)=\frac{1}{2}-\frac{1}{2}cos\left(\frac{14\pi}{5}-2x\right)\)
\(\Leftrightarrow cos\left(8x+\frac{2\pi}{3}\right)=cos\left(2\pi+\frac{4\pi}{5}-2x\right)\)
\(\Leftrightarrow cos\left(8x+\frac{2\pi}{3}\right)=cos\left(\frac{4\pi}{5}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}8x+\frac{2\pi}{3}=\frac{4\pi}{5}-2x+k2\pi\\8x+\frac{2\pi}{3}=2x-\frac{4\pi}{5}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{75}+\frac{k\pi}{5}\\x=-\frac{11\pi}{45}+\frac{k\pi}{3}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2020
a.
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos4x=\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{2\pi}{3}\right)\)
\(\Leftrightarrow cos4x=-cos\left(2x+\frac{2\pi}{3}\right)\)
\(\Leftrightarrow cos4x=cos\left(\frac{\pi}{3}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{3}-2x+k2\pi\\4x=2x-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{18}+\frac{k\pi}{3}\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
b.
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos\left(10x+\frac{2\pi}{3}\right)-\frac{1}{2}-\frac{1}{2}cos\left(6x+\frac{\pi}{2}\right)=0\)
\(\Leftrightarrow cos\left(10x+\frac{2\pi}{3}\right)=-cos\left(6x+\frac{\pi}{2}\right)\)
\(\Leftrightarrow cos\left(10x+\frac{2\pi}{3}\right)=cos\left(\frac{\pi}{2}-6x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}10x+\frac{2\pi}{3}=\frac{\pi}{2}-6x+k2\pi\\10x+\frac{2\pi}{3}=6x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{96}+\frac{k\pi}{8}\\x=-\frac{7\pi}{24}+\frac{k\pi}{2}\end{matrix}\right.\)
4 tháng 6 2019
Các bước biến đổi. Bạn tự tìm kết quả nhé!
1) \(\left(\sin x-\cos x\right)\left(\cos^2x+\cos x.\sin x+\sin^2x\right)+\cos^2x-\sin^2x=0\)
<=> \(\left(\sin x-\cos x\right)\left(1+\cos x.\sin x\right)+\left(\cos x-\sin x\right)\left(\cos x+\sin x\right)=0\)
<=> \(\left(\sin x-\cos x\right)\left(\cos x+1\right)\left(\sin x+1\right)=0\)
2) \(\left(\sin^3x-2\sin^5x\right)-\left(2\cos^5x-\cos^3x\right)=0\)
<=> \(\sin^3x\left(1-2\sin^2x\right)-\cos^3x\left(2\cos^2x-1\right)=0\)
<=> \(\sin^3x.\cos2x-\cos^3x.\cos2x=0\)
<=> \(\cos2x\left(\sin^3x-\cos^3x\right)=0\)
3) ĐK: x\(\ne\frac{\pi}{2}+k\pi\)
\(\cos x\left(3.\tan x+2\right)-\left(3\tan x+2\right)=0\)
<=> \(\left(\cos x-1\right)\left(3.\tan x+2\right)=0\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 9 2020
\(sina+sinb+sinc+3=0\)
\(\Leftrightarrow\left(sina+1\right)+\left(sinb+1\right)+\left(sinc+1\right)=0\)
Do \(\left\{{}\begin{matrix}sina\ge-1\\sinb\ge-1\\sinc\ge-1\end{matrix}\right.\) ;\(\forall a;b;c\)
\(\Rightarrow\left(sina+1\right)+\left(sinb+1\right)+\left(sinc+1\right)\ge0\)
Dấu "=" xảy ra khi và chỉ khi \(sina=sinb=sinc=-1\)
\(\Rightarrow cosa=cosb=cosc=0\Rightarrow cosa+cosb+cosc+10=10\)
b/ \(sinx=1-sin^2x\Rightarrow sinx=cos^2x\)
\(\Rightarrow sin^2x=cos^4x\Rightarrow1-cos^2x=cos^4x\)
\(\Rightarrow cos^4x+cos^2x=1\Rightarrow\left(cos^4x+cos^2x\right)^2=1\)
\(\Rightarrow cos^8x+2cos^6x+cos^4x=1\)
TG
1
1 tháng 9 2020
\(1.sin3x+sin2x+sinx=cos2x+cosx+1\)
\(\Leftrightarrow2sin2x.cosx+sin2x=2cos^2x+cosx\)
\(\Leftrightarrow sin2x\left(2cosx+1\right)-cosx\left(2cosx+1\right)=0\\\)
\(\Leftrightarrow\left(2cosx-1\right)\left(sin2x-cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin2x=sin\left(\frac{\Pi}{2}-x\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\frac{\Pi}{3}+k2\Pi\\x=\frac{\Pi}{6}+m2\Pi orx=\frac{\Pi}{2}+k2\Pi\end{matrix}\right.\)
\(2.cos^2x+cos^23x=sin^22x\)
\(\Leftrightarrow2+cos2x+cos6x=1-cos4x\)
\(\Leftrightarrow1+cos2x+cos6x+cos4x=0\)
\(\Leftrightarrow2cos^2x+2cos5x.cosx=0\)
\(\Leftrightarrow2cosx\left(cosx+cos5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\Pi}{2}+k\Pi\\cos5x=cos\left(\Pi-x\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\Pi}{2}+k\Pi\\5x=\Pi-x+k2\Pi or5x=x-\Pi+k2\Pi\end{matrix}\right.\)
NT
1
HT
21 tháng 8 2020
a/ \(\Leftrightarrow\cos\left(\frac{\pi}{7}-3x\right)=\cos\left(-\frac{5}{6}\pi\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{\pi}{7}-3x=-\frac{5}{6}\pi+k2\pi\\\frac{\pi}{7}-3x=\frac{5}{6}\pi+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{41}{126}\pi-\frac{2}{3}k\pi\\x=-\frac{29}{42}\pi-\frac{2}{3}k\pi\end{matrix}\right.\)
b/ \(\Leftrightarrow\sin\left(90^0-\frac{x}{3}\right)=\sin\left(2x+30^0\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}90^0-\frac{x}{3}=2x+30^0+k180^0\\90^0-\frac{x}{3}=180^0-2x-30^0+k180^0\end{matrix}\right.\Leftrightarrow...\)
c/ \(DKXD:\cos\left(30^0-2x\right)\ne0\Leftrightarrow30^0-2x\ne90^0+k180^0\Leftrightarrow x\ne-30^0-k90^0\)
\(\Leftrightarrow30^0-2x=60^0+k180^0\Leftrightarrow x=-15^0-k90^0\left(tm\right)\)
d/ \(DKXD:\sin\left(30^0-2x\right)\ne0\Leftrightarrow30^0-2x\ne k180^0\Leftrightarrow x\ne15^0-k90^0\)
\(\Leftrightarrow30^0-2x=30^0+k.180^0\Leftrightarrow x=-k.90^0\left(tm\right)\)
\(\Leftrightarrow sin^25x+sin^23x=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin3x=0\\sin5x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)
\(\Leftrightarrow x=k\pi\)