\(2sin^2x+\sqrt{3}sin2x-2=0\)
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NP
Nguyễn Phương Thảo
28 tháng 7 2020
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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
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x=3\)
\(\Leftrightarrow\sqrt{3}sin2x-cos2x=2\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x=1\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{6}\right)=1\)
\(\Leftrightarrow2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\)
NT
1
19 tháng 7 2022
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}\cdot\cos2x+\dfrac{1}{2}\cdot\sin2x+\sin\left(2x+\dfrac{\Pi}{6}\right)=\sqrt{2}\)
\(\Leftrightarrow\sin\left(2x+\dfrac{\Pi}{3}\right)+\sin\left(2x+\dfrac{\Pi}{6}\right)=\sqrt{2}\)
\(\Leftrightarrow2\cdot\dfrac{\sin\left(2x+\dfrac{\Pi}{3}+2x+\dfrac{\Pi}{6}\right)}{2}\cdot\dfrac{\cos\left(2x+\dfrac{\Pi}{3}-2x-\dfrac{\Pi}{6}\right)}{2}=\sqrt{2}\)
\(\Leftrightarrow\sin\left(4x+\dfrac{\Pi}{2}\right)\cdot\cos\left(\dfrac{\Pi}{6}\right)=2\sqrt{2}\)
\(\Leftrightarrow\sin\left(4x+\dfrac{\Pi}{2}\right)=\dfrac{4\sqrt{6}}{3}\)
hay \(x\in\varnothing\)
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/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=...\)
2 tháng 8 2022
a: \(A=\sqrt{3}\left(\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx\right)+\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\)
\(=\dfrac{\sqrt{3}}{2}sinx-\dfrac{3}{2}cosx+\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\)
\(=\sqrt{3}sinx-cosx\)
c: \(=2\left[\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x\right]+4sinx+1\)
\(=\sqrt{3}sin2x-cos2x+4sinx+1\)
d: \(D=\sqrt{3}cos2x+sin2x+2\cdot\left(\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x\right)\)
\(=\sqrt{3}\cdot cos2x+sin2x+\sqrt{3}\cdot sin2x-cos2x\)
\(=cos2x\left(\sqrt{3}-1\right)+sin2x\left(1+\sqrt{3}\right)\)
MT
1
13 tháng 6 2023
a: =>sin2x+2*(1-cos2x)/2=2
=>sin2x-cos2x=1
=>căn 2*sin(2x-pi/4)=1
=>2x-pi/4=pi/4+k2pi hoặc 2x-pi/4=3/4pi+k2pi
=>x=pi/4+kpi hoặc x=pi/2+kpi
b: =>2*(1+cos2x)/2+1/2*sin2x-1/2(1-cos2x)=0
=>1+cos2x+1/2*sin2x-1/2+1/2cos2x=0
=>1/2*sin2x+3/2*cos2x=-1/2
=>sin(2x+a)=-cos(a)=cos(pi-a)
=>sin(2x+a)=sin(-pi/2+a)
=>2x+a=-pi/2+a+k2pi hoặc 2x+a=3/2pi-a+k2pi
=>x=-pi/4+kpi hoặc x=3/4pi-a+kpi
5 tháng 9 2020
ngại viết quá hihi, mà hơi ngáo tí cái dạng này lm rồi mà cứ quên
bài trước mk bình luận bạn đọc chưa nhỉ
NH
\(\dfrac{2sin^3x+2\sqrt{3}sin^2x.cosx-2sin^2x+cos\left(2x+\dfrac{\pi}{3}\right)}{2cosx-\sqrt{3}}=0\)
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