sin2x(cotx+tan2x)=4cos2x
giải hộ mình vs pls :((
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S
2
23 tháng 7 2021
(sin2x - 4cos2x)(sin2x - 2sinx.cosx) = 2cos4x
⇔ (5sin2x - 4)(sin2x - sin2x) = 2cos4x
⇔ \(\left(\dfrac{5-5cos2x}{2}-4\right)\left(\dfrac{1-cos2x}{2}-sin2x\right)\)= 2cos4x
⇔ \(\dfrac{5-5cos2x-8}{2}.\dfrac{1-cos2x-2sin2x}{2}\) = 2cos4x
⇔ (5cos2x + 3)(cos2x + 2sin2x - 1) = 8cos4x
⇔ 5cos22x + 5cos2x.sin2x + 3cos2x + 6sin2x - 3 = 8cos4x
⇔ 5.\(\dfrac{1+cos4x}{2}\) + \(\dfrac{5}{2}sin4x\) + 3cos2x + 6sin2x - 3 = 8cos4x
⇔ \(\dfrac{5}{2}cos4x+\dfrac{5}{2}sin4x+3cos2x+6sin2x-\dfrac{1}{2}\) = 8cos4x
⇔ 5cos4x + 5sin4x + 6cos2x + 12sin2x - 1 = 16cos4x
VP = 16cos4x = 16 . \(\dfrac{\left(1+cos2x\right)^2}{4}\) = 4. (1 + cos2x)2
VP = 4 . (1 + 2cos2x + cos22x)
VP = 4 + 8cos2x + 4 . \(\dfrac{1+cos4x}{2}\)
VP = 6 + 8cos2x+ 2cos4x
Vậy 3cos4x + 5sin4x - 2cos2x + 12sin2x - 7 = 0
N
1
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 9 2021
1.
\(tan^2x-5tanx+6=0\)
\(\Rightarrow\left[{}\begin{matrix}tanx=2\\tanx=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(2\right)+k\pi\\x=arctan\left(3\right)+k\pi\end{matrix}\right.\)
2.
\(3cos^22x+4cos2x+1=0\)
\(\Rightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\pi+k2\pi\\2x=\pm arccos\left(-\dfrac{1}{3}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pm\dfrac{1}{2}arccos\left(-\dfrac{1}{3}\right)+k\pi\end{matrix}\right.\)
TH
1
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 11 2019
ĐKXĐ: ...
a/ \(\frac{sin2x}{cos2x}+\frac{cosx}{sinx}=8cos^2x\)
\(\Leftrightarrow sin2x.sinx+cos2x.cosx=8cos^2x.sinx.cos2x\)
\(\Leftrightarrow cosx=4sin2x.cos2x.cosx\)
\(\Leftrightarrow cosx=2sin4x.cosx\)
\(\Leftrightarrow cosx\left(2sin4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin4x=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)
b/ \(\frac{cosx}{sinx}-\frac{sinx}{cosx}+4sin2x=\frac{1}{sinx.cosx}\)
\(\Leftrightarrow cos^2x-sin^2x+4sin2x.sinx.cosx=1\)
\(\Leftrightarrow cos2x+2sin^22x=1\)
\(\Leftrightarrow cos2x+2\left(1-cos^22x\right)=1\)
\(\Leftrightarrow-2cos^22x+cos2x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=1\\cos2x=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 11 2019
1c/
\(5sinx-2=3\left(1-sinx\right)\frac{sin^2x}{1-sin^2x}\)
\(\Leftrightarrow5sinx-2=\frac{3sin^2x}{1+sinx}\)
\(\Leftrightarrow\left(5sinx-2\right)\left(1+sinx\right)=3sin^2x\)
\(\Leftrightarrow5sin^2x+3sinx-2=3sin^2x\)
\(\Leftrightarrow2sin^2x+3sinx-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=-2\left(l\right)\end{matrix}\right.\) \(\Rightarrow x=...\)
Bài 2:
a/ \(\Leftrightarrow\frac{\left(m+1\right)\left(1-cos2x\right)}{2}-sin2x+cos2x=0\)
\(\Leftrightarrow2sin2x+\left(m-1\right)cos2x=m+1\)
Theo điều kiện có nghiệm của pt lượng giác bậc nhất:
\(4+\left(m-1\right)^2\ge\left(m+1\right)^2\)
\(\Leftrightarrow4m\le4\Rightarrow m\le1\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 7 2020
c/
\(\Leftrightarrow2sinx.cosx-2\sqrt{3}cos^2x=0\)
\(\Leftrightarrow2cosx\left(sinx-\sqrt{3}cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx-\sqrt{3}cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\sinx=\sqrt{3}cosx\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\frac{sinx}{cosx}=\sqrt{3}\Leftrightarrow tanx=\sqrt{3}\)
\(\Rightarrow x=\frac{\pi}{3}+k\pi\)
d/
\(\Leftrightarrow tan\left(3x-50^0\right)=-cot\left(x-30^0\right)\)
\(\Leftrightarrow tan\left(3x-50^0\right)=tan\left(x+60^0\right)\)
\(\Rightarrow3x-50^0=x+60^0+k180^0\)
\(\Rightarrow x=55^0+k90^0\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 7 2020
a/
\(\Leftrightarrow sinx=2cosx\)
Nhận thấy \(cosx=0\) không phải nghiệm, pt tương đương:
\(\frac{sinx}{cosx}=2\Leftrightarrow tanx=2\)
\(\Leftrightarrow tanx=tana\) (với \(a\in\left(0;\frac{\pi}{2}\right)\) sao cho \(tana=2\))
\(\Rightarrow x=a+k\pi\)
b/
\(tan2x=cotx=tan\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow2x=\frac{\pi}{2}-x+k\pi\)
\(\Rightarrow x=\frac{\pi}{6}+\frac{k\pi}{3}\)
16 tháng 10 2023
a: \(VT=\dfrac{cot^2x}{1+cot^2x}\cdot\dfrac{1+tan^2x}{tan^2x}\)
\(=\dfrac{cot^2x}{\dfrac{1}{sin^2x}}\cdot\dfrac{\dfrac{1}{cos^2x}}{tan^2x}\)
\(=\dfrac{cot^2x}{tan^2x}\cdot\dfrac{1}{cos^2x}:\dfrac{1}{sin^2x}\)
\(=\dfrac{cot^2x}{tan^2x}\cdot\dfrac{sin^2x}{cos^2x}\)
\(=cot^2x\)
\(VP=\dfrac{tan^2x+cot^2x}{1+tan^4x}=\dfrac{\dfrac{sin^2x}{cos^2x}+\dfrac{cos^2x}{sin^2x}}{1+\dfrac{sin^4x}{cos^4x}}\)
\(=\dfrac{sin^4x+cos^4x}{sin^2x\cdot cos^2x}:\dfrac{cos^4x+sin^4x}{cos^4x}\)
\(=\dfrac{sin^4x+cos^4x}{sin^2x\cdot cos^2x}\cdot\dfrac{cos^4x}{cos^4x+sin^4x}=\dfrac{cos^2x}{sin^2x}=cot^2x\)
=>VT=VP
b:
\(\dfrac{tan^2x-cos^2x}{sin^2x}+\dfrac{cot^2x-sin^2x}{cos^2x}\)
\(=\dfrac{\left(\dfrac{sinx}{cosx}\right)^2-cos^2x}{sin^2x}+\dfrac{\left(\dfrac{cosx}{sinx}\right)^2-sin^2x}{cos^2x}\)
\(=\dfrac{sin^2x-cos^4x}{cos^2x\cdot sin^2x}+\dfrac{cos^2x-sin^4x}{sin^2x\cdot cos^2x}\)
\(=\dfrac{sin^2x+cos^2x-cos^4x-sin^4x}{cos^2x\cdot sin^2x}\)
\(=\dfrac{1-\left(cos^2x+sin^2x\right)^2+2\cdot cos^2x\cdot sin^2x}{cos^2x\cdot sin^2x}\)
\(=\dfrac{2\cdot cos^2x\cdot sin^2x}{cos^2x\cdot sin^2x}=2\)
