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NA
2
17 tháng 8 2020
\(\left(2\sin x-1\right)\left(2\sin2x+1\right)=3-4\cos^2x\)
\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=3-4\left(2-\sin^2x\right)\)
\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=4\sin^2x-1\)
\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=\left(2\sin x-1\right)\left(2\sin x+1\right)\)
\(\Leftrightarrow2\sin2x+1=2\sin x+1\)
\(\Leftrightarrow\sin2x=\sin x\)
\(\Leftrightarrow\sin2x-\sin x=0\)
\(\Leftrightarrow2\cos\frac{3}{2}-\cos\frac{x}{2}=0\)
\(\Leftrightarrow\orbr{\begin{cases}\cos\frac{3}{2}=0\\\cos\frac{x}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x}{2}=\frac{\pi}{2}+k2\pi\\\frac{x}{2}=\frac{\pi}{2}+k2\pi\end{cases}\left(k\inℤ\right)}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\pi}{3}+\frac{2\pi}{3}k\\x=\pi+4k\pi\end{cases}\left(k\inℤ\right)}\)
TA
17 tháng 7 2021
`2sin^2x+\sqrt3sin2x=3`
`<=>2. (1-cos2x)/2 + \sqrt3sin2x=3`
`<=>\sqrt3sin2x-cos2x=2`
`<=> \sqrt3/2 sin2x-1/2 cos2x=1`
`<=>sin (2x-π/6) = 1`
`<=> 2x-π/6=π/2+k2π`
`<=> x=π/3+kπ (k \in ZZ)`.
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 7 2021
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x=3\)
\(\Leftrightarrow\sqrt{3}sin2x-cos2x=2\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x=1\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{6}\right)=1\)
\(\Leftrightarrow2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 12 2020
\(\Leftrightarrow2sinx.cosx-2cosx+2sin^2x+sinx-3=0\)
\(\Leftrightarrow2cosx\left(sinx-1\right)+\left(sinx-1\right)\left(2sinx+3\right)=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2cosx+2sinx+3\right)=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{3}{2\sqrt{2}}\end{matrix}\right.\)
\(\Leftrightarrow...\)
NS
2
30 tháng 6 2021
Pt \(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{4}=-\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{\pi}{4}=\dfrac{4\pi}{3}+k2\pi\end{matrix}\right.\),\(k\in Z\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{24}+k\pi\\x=\dfrac{19\pi}{24}+k\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy...
Hôm qua họ bảo toi ra lấy CCCD nma toi chưa đi, nay toi đi họ lại đang họp, liệu mai toi đi có bị ăn chửi ko, mn cho ý kiến đi :<
HP
30 tháng 6 2021
\(2sin\left(2x-\dfrac{\pi}{4}\right)+\sqrt{3}=0\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)=sin\left(-\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{4}=-\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{\pi}{4}=\pi+\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{12}+k2\pi\\2x=\dfrac{19\pi}{12}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{24}+k\pi\\x=\dfrac{19\pi}{24}+k\pi\end{matrix}\right.\)
21 tháng 9 2023
b) \(2sin^2x-3sinxcosx+cos^2x=0\)
\(\Leftrightarrow2tan^2x-3tanx+1=0\left(cosx\ne0\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=tan\dfrac{\pi}{4}\\tanx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=arctan\left(\dfrac{1}{2}\right)+k\pi\end{matrix}\right.\left(k\in Z\right)\)
(k ∈ Z)
28 tháng 5 2021
\(\Leftrightarrow1-cos4x+sin7x-1=sinx\)
\(\Leftrightarrow sin7x-sinx-cos4x=0\)
\(\Leftrightarrow2.cos4x.sin3x-cos4x=0\)
\(\Leftrightarrow cos4x\left(2.sin3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\sin3x=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{2}+k\pi\\3x=\dfrac{\pi}{6}+k2\pi\\3x=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)(\(k\in Z\)) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\) (\(k\in Z\))
Kết luận:...
\(\Leftrightarrow2\left(1-cos^22x\right)=2+\left(1-2sin^2x\right)\)
\(\Leftrightarrow2-2cos^22x=2+cos2x\)
\(\Leftrightarrow2cos^22x+cos2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\2x=\pm\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=\pm\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)
số 3 ban đầu đã đi đâu v