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TM
1
2
13 tháng 9 2023
`cos 2x+\sqrt{3}sin 2x+\sqrt{3}sin x-cos x=4`
`<=>1/2 cos 2x+\sqrt{3}/2 sin 2x+\sqrt{3}/2 sin x-1/2 cos x=2`
`<=>sin(\pi/6 +2x)+sin(x-\pi/6)=2`
Vì `-1 <= sin (\pi/6 +2x) <= 1`
`-1 <= sin (x-\pi/6) <= 1`
Dấu "`=`" xảy ra `<=>{(sin(\pi/6+2x)=1),(sin(x-\pi/6)=1):}`
`<=>{(\pi/6+2x=\pi/2+k2\pi),(x-\pi/6=\pi/2+k2\pi):}`
`<=>{(x=\pi/6+k\pi),(x=[2\pi]/3+k2\pi):}` `(k in ZZ)`
PD
0
MA
0
NH
2
HP
20 tháng 8 2021
\(\sqrt{2}sinx+sin2x=\sqrt{3}cos2x-\sqrt{6}cosx\)
\(\Leftrightarrow\dfrac{\sqrt{2}}{2}sinx+\dfrac{1}{2}sin2x-\dfrac{\sqrt{3}}{2}cos2x+\dfrac{\sqrt{6}}{2}cosx=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{3}\right)+sin\left(2x-\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow\sqrt{2}cos\left(x-\dfrac{\pi}{6}\right)+2sin\left(x-\dfrac{\pi}{6}\right).cos\left(x-\dfrac{\pi}{6}\right)=0\)
\(\Leftrightarrow\sqrt{2}cos\left(x-\dfrac{\pi}{6}\right)\left[1+\sqrt{2}sin\left(x-\dfrac{\pi}{6}\right)\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{6}\right)=0\\sin\left(x-\dfrac{\pi}{6}\right)=-\dfrac{1}{\sqrt{2}}\end{matrix}\right.\)
Đến đấy thì dễ rồi.
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 8 2021
\(\Leftrightarrow\sqrt{2}\left(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx\right)+\dfrac{1}{2}sin2x-\dfrac{\sqrt{3}}{2}cos2x=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{3}\right)+sin\left(2x-\dfrac{\pi}{3}\right)=0\)
Đặt \(x+\dfrac{\pi}{3}=u\Rightarrow2x-\dfrac{\pi}{3}=2u-\pi\)
\(\Rightarrow\sqrt{2}sinu+sin\left(2u-\pi\right)=0\)
\(\Leftrightarrow\sqrt{2}sinu-sin2u=0\)
\(\Leftrightarrow sinu\left(\sqrt{2}-2cosu\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinu=0\\cosu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{3}\right)=0\\cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 10 2020
1.
\(\Leftrightarrow sin^2x\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cos^2x\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(1+cosx\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(sinx+cosx+sinx.cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\Leftrightarrow...\\sinx+cosx+sinx.cosx-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
Đặt \(sinx+cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow t+\frac{t^2-1}{2}-1=0\)
\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 10 2020
2.
\(\Leftrightarrow\sqrt{3}sinx.cosx+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)
\(\Leftrightarrow cosx\left(\sqrt{3}sinx+\sqrt{2}cosx+\sqrt{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Leftrightarrow...\\\sqrt{3}sinx+\sqrt{2}cosx=-\sqrt{6}\left(1\right)\end{matrix}\right.\)
Xét (1):
Do \(\sqrt{3}^2+\sqrt{2}^2< \left(-\sqrt{6}\right)^2\) nên (1) vô nghiệm
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 9 2020
3.
\(4sinx.cosx-2sinx+1-2cosx=0\)
\(\Leftrightarrow2sinx\left(2cosx-1\right)-\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
4.
\(cosx-sinx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\-4sinx.cosx=2t^2-2\end{matrix}\right.\)
Pt trở thành: \(t+2t^2-2-1=0\Leftrightarrow2t^2+t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-\frac{3}{2}< -\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}cos\left(x+\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\\x+\frac{\pi}{4}=-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 9 2020
5.
\(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x=sinx\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=x+k2\pi\\2x+\frac{\pi}{6}=\pi-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
6.
\(9sin^2x-5\left(1-sin^2x\right)-5sinx+4=0\)
\(\Leftrightarrow14sin^2x-5sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=-\frac{1}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=arcsin\left(-\frac{1}{7}\right)+k2\pi\\x=\pi-arcsin\left(-\frac{1}{7}\right)+k2\pi\end{matrix}\right.\)