CỨU MÌNH VS
a) 3Cos^2x -2Sin2x + Sin^2x =1
b) 4Cos^2x -3SinxCosx +3Sin^2x =1
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HP
3 tháng 9 2021
1.
\(3sin^22x-2sin2x.cos2x-4cos^22x=2\)
\(\Leftrightarrow-\dfrac{3}{2}\left(1-2sin^22x\right)-2sin2x.cos2x-2\left(2cos^22x-1\right)=\dfrac{5}{2}\)
\(\Leftrightarrow sin4x+\dfrac{7}{2}cos4x=-\dfrac{5}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{53}}{2}\left(\dfrac{2}{\sqrt{53}}sin4x+\dfrac{7}{\sqrt{53}}cos4x\right)=-\dfrac{5}{2}\)
\(\Leftrightarrow sin\left(4x+arccos\dfrac{2}{\sqrt{53}}\right)=-\dfrac{5}{\sqrt{53}}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+arccos\dfrac{2}{\sqrt{53}}=arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+k2\pi\\4x+arccos\dfrac{2}{\sqrt{53}}=\pi-arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}arccos\dfrac{2}{\sqrt{53}}+\dfrac{1}{4}arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}-\dfrac{1}{4}arccos\dfrac{2}{\sqrt{53}}-\dfrac{1}{4}arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+\dfrac{k\pi}{2}\end{matrix}\right.\)
HP
3 tháng 9 2021
2.
\(2\sqrt{3}cos^2x+6sinx.cosx=3+\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}\left(2cos^2x-1\right)+6sinx.cosx=3\)
\(\Leftrightarrow\sqrt{3}cos2x+3sin2x=3\)
\(\Leftrightarrow2\sqrt{3}\left(\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x\right)=3\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\2x-\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 6 2020
\(B=cos^2x.cot^2x+cos^2x-cot^2x+2\left(sin^2x+cos^2x\right)\)
\(=cos^2x\left(cot^2x+1\right)-cot^2x+2\)
\(=\frac{cos^2x}{sin^2x}-cot^2x+1=cot^2x-cot^2x+1=1\)
\(M=cos^4x-sin^4x+cos^4x+sin^2x.cos^2x+3sin^2x\)
\(=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+cos^2x\left(cos^2x+sin^2x\right)+3sin^2x\)
\(=cos^2x-sin^2x+cos^2x+3sin^2x\)
\(=2\left(sin^2x+cos^2x\right)=2\)
21 tháng 9 2023
b) \(2sin^2x-3sinxcosx+cos^2x=0\)
\(\Leftrightarrow2tan^2x-3tanx+1=0\left(cosx\ne0\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=tan\dfrac{\pi}{4}\\tanx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=arctan\left(\dfrac{1}{2}\right)+k\pi\end{matrix}\right.\left(k\in Z\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 12 2021
\(P=\sqrt{\left(1-cos^2x\right)^2+6cos^2x+3cos^4x}+\sqrt{\left(1-sin^2x\right)^2+6sin^2x+3sin^4x}\)
\(=\sqrt{4cos^4x+4cos^2x+1}+\sqrt{4sin^4x+4sin^2x+1}\)
\(=\sqrt{\left(2cos^2x+1\right)^2}+\sqrt{\left(2sin^2x+1\right)^2}\)
\(=2cos^2x+1+2sin^2x+1\)
\(=2\left(sin^2x+cos^2x\right)+2=4\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 10 2020
a/
Nhận thấy \(cosx=0\) không phải nghiệm, chia 2 vế cho \(cos^2x\)
\(\Leftrightarrow3tan^2x+8tanx+8\sqrt{3}-9=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-\sqrt{3}\\tanx=\frac{3\sqrt{3}-8}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{3}+k\pi\\x=arctan\left(\frac{3\sqrt{3}-8}{3}\right)+k\pi\end{matrix}\right.\)
b/
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)
\(tan^2x+2tanx-2=\frac{1}{2}\left(1+tan^2x\right)\)
\(\Leftrightarrow tan^2x+4tanx-5=0\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=arctan\left(-5\right)+k\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\left(sinx+1\right)\left(1-2sin^2x-1\right)=0\)
\(\Leftrightarrow sin^2x\left(sinx+1\right)=0\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 8 2021
a.
\(y=2\left(1-cos2x\right)-\dfrac{5}{2}sin2x+\dfrac{1}{2}+\dfrac{1}{2}cos2x+10\)
\(=-\dfrac{1}{2}\left(5sin2x+3cos2x\right)+\dfrac{25}{2}\)
\(=-\dfrac{\sqrt{34}}{2}\left(\dfrac{5}{\sqrt{34}}sin2x+\dfrac{3}{\sqrt{34}}cos2x\right)+\dfrac{25}{2}\)
Đặt \(\dfrac{5}{\sqrt{34}}=cosa\)
\(\Rightarrow y=-\dfrac{\sqrt{34}}{2}\left(sin2x.cosa+cos2x.sina\right)+\dfrac{25}{2}\)
\(=-\dfrac{\sqrt{34}}{2}sin\left(2x+a\right)+\dfrac{25}{2}\)
Do \(-1\le sin\left(2x+a\right)\le1\)
\(\Rightarrow\dfrac{25-\sqrt{34}}{2}\le y\le\dfrac{25+\sqrt{34}}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 8 2021
b.
\(y=\dfrac{sin^2x-2sin2x+1}{3+sin^2x+2cos^2x}=\dfrac{2sin^2x-4sin2x+2}{6+2\left(sin^2x+cos^2x\right)+2cos^2x}\)
\(=\dfrac{1-cos2x-4sin2x+2}{8+1+cos2x}=\dfrac{3-4sin2x-cos2x}{9+cos2x}\)
\(\Rightarrow9y+y.cos2x=3-4sin2x-cos2x\)
\(\Rightarrow4sin2x+\left(y+1\right)cos2x=3-9y\)
Theo điều kiện có nghiệm của pt lượng giác bậc nhất:
\(4^2+\left(y+1\right)^2\ge\left(3-9y\right)^2\)
\(\Leftrightarrow80y^2-56y-8\le0\)
\(\Rightarrow\dfrac{7-\sqrt{89}}{20}\le y\le\dfrac{7+\sqrt{89}}{20}\)
a/
\(\Leftrightarrow3cos^2x-4sinx.cosx+1-cos^2x=1\)
\(\Leftrightarrow2cos^2x-4sinx.cosx=0\)
\(\Leftrightarrow2cosx\left(cosx-2sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\tanx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=arctan\left(\frac{1}{2}\right)+k\pi\end{matrix}\right.\)
b.
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)
\(4-3tanx+3tan^2x=1+tan^2x\)
\(\Leftrightarrow2tan^2x-3tanx+3=0\)
Pt vô nghiệm