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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
31 tháng 7 2020
e/
ĐKXĐ: ...
\(\Leftrightarrow\frac{2sin4x.cos2x}{cos2x}-2cos4x=2\sqrt{2}\)
\(\Leftrightarrow2sin4x-2cos4x=2\sqrt{2}\)
\(\Leftrightarrow sin4x-cos4x=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(4x-\frac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sin\left(4x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow4x-\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\frac{3\pi}{16}+\frac{k\pi}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 7 2020
d/
Đặt \(sin2x-cos2x=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=t\Rightarrow\left|t\right|\le\sqrt{2}\)
\(\Rightarrow t^2-3t-4=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\2x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{3\pi}{4}+k\pi\end{matrix}\right.\)
O
0
MM
2
HP
28 tháng 9 2021
a, \(sin4x.cosx-sin3x=0\)
\(\Leftrightarrow\dfrac{1}{2}sin5x+\dfrac{1}{2}sin3x-sin3x=0\)
\(\Leftrightarrow sin5x=sin3x\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=3x+k2\pi\\5x=\pi-3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\end{matrix}\right.\)
HP
28 tháng 9 2021
b, \(sin2x+\sqrt{3}cos2x=\sqrt{2}\)
\(\Leftrightarrow\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}cos2x=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{3}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=\dfrac{\pi}{4}+k2\pi\\2x+\dfrac{\pi}{3}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{24}+k\pi\\x=\dfrac{5\pi}{24}+k\pi\end{matrix}\right.\)
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)`
NL
0
PD
1
18 tháng 4 2016
Từ phương trình ban đầu ta có : \(2\cos5x\sin x=\sqrt{3}\sin^2x+\sin x\cos x\)
\(\Leftrightarrow\begin{cases}\sin x=0\\2\cos5x=\sqrt{3}\sin x+\cos x\end{cases}\)
+) \(\sin x=0\Leftrightarrow x=k\pi\)
+)\(2\cos5x=\sqrt{3}\sin x+\cos x\Leftrightarrow\cos5x=\cos\left(x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\begin{cases}x=-\frac{\pi}{12}+\frac{k\pi}{2}\\x=\frac{\pi}{18}+\frac{k\pi}{3}\end{cases}\)
TC
GPT: \(\dfrac{\left(\sin x-\cos x\right)\left(\sin2x-3\right)-\sin2x-\cos2x+1}{2\sin x-\sqrt{2}}=0\)
1
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 1 2021
ĐKXĐ: \(sinx\ne\dfrac{\sqrt{2}}{2}\)
\(\left(sinx-cosx\right)\left(sin2x-3\right)+\left(sinx-cosx\right)^2+\left(sin^2x-cos^2x\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(sin2x-3\right)+\left(sinx-cosx\right)^2+\left(sinx-cosx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(sin2x-3+2sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\\left(sin2x-1\right)+2\left(sinx+1\right)=0\left(vô-nghiệm\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)
Kết hợp ĐKXĐ \(\Rightarrow x=-\dfrac{\pi}{4}+k2\pi\)
17 tháng 7 2021
Pt \(\Leftrightarrow2sin\left(2x+\dfrac{\pi}{3}\right)=\sqrt{3}\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=k\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(x\in\left(0;\dfrac{\pi}{2}\right)\)\(\Rightarrow\left[{}\begin{matrix}0< \dfrac{\pi}{6}+k\pi< \dfrac{\pi}{2}\\0< k\pi< \dfrac{\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{1}{6}< k< \dfrac{1}{3}\\0< k< \dfrac{1}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Leftrightarrow\left[{}\begin{matrix}k=0\\k\in\varnothing\end{matrix}\right.\)
Vậy có 1 nghiệm thỏa mãn
\(\sqrt{3}sin2x-cos2x=\sqrt{2}\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow cos\left(\dfrac{\pi}{6}\right)sin2x-sin\left(\dfrac{\pi}{6}\right)cos2x=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{4}\right)\)
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