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NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 8 2020
\(\Leftrightarrow4sin^23x.sin^2x=6+2sin3x\)
Do \(\left\{{}\begin{matrix}sin^23x\le1\\sin^2x\le1\end{matrix}\right.\) \(\Rightarrow VT\le4\)
\(sin3x\ge-1\Rightarrow VP=6+2sin3x\ge4\)
\(\Rightarrow VP\ge VT\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}sin^23x=1\\sin^2x=1\\sin3x=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sin^2x=1\\sin3x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin^2x=1\\3sinx-4sin^3x=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sin^2x=1\\sinx\left(3-4sin^2x\right)=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin^2x=1\\sinx=1\end{matrix}\right.\) \(\Leftrightarrow sinx=1\)
\(\Rightarrow x=\frac{\pi}{2}+k2\pi\)
3 tháng 7 2021
a) Pt \(\Leftrightarrow3.cos4x-\left(cos6x+1\right)=1\)
\(\Leftrightarrow3cos4x-cos6x-2=0\)
Đặt \(t=2x\)
Pttt:\(3cos2t-cos3t-2=0\)
\(\Leftrightarrow3\left(2cos^2t-1\right)-\left(4cos^3t-3cost\right)-2=0\)
\(\Leftrightarrow-4cos^3t+6cos^2t+3cost-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cost=1\\cost=\dfrac{1+\sqrt{21}}{4}\left(vn\right)\\cost=\dfrac{1-\sqrt{21}}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=k2\pi\\t=\pm arc.cos\left(\dfrac{1-\sqrt{21}}{4}\right)+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{1}{2}.arccos\left(\dfrac{1-\sqrt{21}}{4}\right)+k\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
a2) \(2cos2x-8cosx+7=\dfrac{1}{cosx}\) (ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\))
\(\Leftrightarrow2.\left(2cos^2x-1\right)-8cosx+7=\dfrac{1}{cosx}\)
\(\Leftrightarrow2.\left(2cos^2x-1\right)cosx-8cos^2x+7cosx=1\)
\(\Leftrightarrow4cos^3x-8cos^2x+5cosx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) (tm) (\(k\in Z\))
Vậy...
a3) Đk: \(x\ne-\dfrac{\pi}{4}+k\pi;x\ne\dfrac{\pi}{2}+k\pi\)
Pt \(\Leftrightarrow\dfrac{\left(1+sinx+1-2sin^2x\right).\dfrac{1}{\sqrt{2}}\left(sinx+cosx\right)}{1+\dfrac{sinx}{cosx}}=\dfrac{1}{\sqrt{2}}cosx\)
\(\Leftrightarrow\dfrac{\left(-2sin^2x+sinx+2\right).\left(sinx+cosx\right)cosx}{cosx+sinx}=cosx\)
\(\Leftrightarrow\left(2+sinx-2sin^2x\right).cosx=cosx\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\2+sinx-2sin^2x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
3 tháng 7 2021
a4) Pt \(\Leftrightarrow9sinx+6cosx-6sinx.cosx+1-2sin^2x=8\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sin^2x-9sinx+7\right)=0\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sinx-7\right)\left(sinx-1\right)=0\)
\(\Leftrightarrow\left(1-sinx\right)\left(6cosx+2sinx+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\6cosx+2sinx=7\left(vn\right)\end{matrix}\right.\) (\(6cosx+2sinx=7\) vô nghiệm do \(6^2+2^2< 7^2\))
\(\Rightarrow sinx=1\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi;k\in Z\)
Vậy...
A
1
HP
15 tháng 8 2021
ĐK: \(x\ne\dfrac{\pi}{4}+k\pi;x\ne\dfrac{k\pi}{2}\)
\(\dfrac{2sin^2x+cos4x-cos2x}{\left(sinx-cosx\right)sin2x}=0\)
\(\Leftrightarrow2sin^2x+cos4x-cos2x=0\)
\(\Leftrightarrow2sin^2x-1+cos4x-cos2x+1=0\)
\(\Leftrightarrow2cos^22x-2cos2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\2x=k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=k\pi\end{matrix}\right.\)
Đối chiếu điều kiện ta được \(x=-\dfrac{\pi}{4}+k\pi\)
1 tháng 10 2021
a
\(\Leftrightarrow\left(3sinx-sin3x\right)cos3x+\left(3cosx+cos3x\right)sin3x+3\sqrt{3}cos4x=3\)
\(\Leftrightarrow\left(sinx.cos3x+sin3x.cosx\right)+\sqrt{3}cos4x=1\)
\(\Leftrightarrow sin4x+\sqrt{3}cos4x=1\)
Tới đây thôi, mình lười ghi rồi =))
b
\(\Leftrightarrow\left(1-cos2x\right)\left(2sin^2x-1\right)\left(2sin^2+1\right)=cos2x\left(7cos^22x+3cos2x-4\right)\)
\(\Leftrightarrow\left(1-cos2x\right)\left(-cos2x\right)\left(2-cos2x\right)=cos2x\left(7cos^22x+3cos2x+4\right)\)
\(\Leftrightarrow-cos^22x+3cos2x-2=7cos^22x+3cos2x+4\)
\(\Leftrightarrow4cos^22x+3=0\)
=> pt vô nghiệm
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2020
d/
ĐKXĐ: ...
\(\Leftrightarrow tanx-1+cos2x=0\)
\(\Leftrightarrow\frac{sinx}{cosx}-1-\left(sin^2x-cos^2x\right)=0\)
\(\Leftrightarrow\frac{sinx-cosx}{cosx}-\left(sinx-cosx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(\frac{1}{cosx}-sinx-cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\left(1\right)\\\frac{1}{cosx}-sinx-cosx=0\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=0\)
\(\Rightarrow x-\frac{\pi}{4}=k\pi\Rightarrow x=\frac{\pi}{4}+k\pi\)
\(\left(2\right)\Leftrightarrow1-sinx.cosx-cos^2x=0\)
\(\Leftrightarrow sin^2x-sinx.cosx=0\)
\(\Leftrightarrow sinx\left(sinx-cosx\right)=0\)
\(\Leftrightarrow sinx=0\Rightarrow x=k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2020
c/
\(\Leftrightarrow sinx.cos2x-sinx+1-cos2x=0\)
\(\Leftrightarrow sinx\left(cos2x-1\right)-\left(cos2x-1\right)=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\cos2x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\2x=k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=k\pi\end{matrix}\right.\)
13 tháng 9 2021
\(\dfrac{cos4x}{cos2x}=tan2x\). ĐKXĐ : \(x\ne\dfrac{\pi}{4}+k.\dfrac{\pi}{2}\), k là số nguyên (tức là sin2x khác 1 và -1)
⇒ cos4x = sin2x
⇔ 1 - 2sin22x = sin2x
⇔ 2sin22x + sin2x - 1 = 0
⇔ \(\left[{}\begin{matrix}sin2x=-1\left(/\right)\\sin2x=\dfrac{1}{2}\left(V\right)\end{matrix}\right.\)
Mà x ∈ \(\left(0;\dfrac{\pi}{2}\right)\)
⇒ \(\left[{}\begin{matrix}x=\dfrac{\pi}{6}\\x=\dfrac{\pi}{3}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 7 2020
\(cos3x=-cos\left(x-120^0\right)\)
\(\Leftrightarrow cos3x=cos\left(x+60^0\right)\)
\(\Rightarrow\left[{}\begin{matrix}3x=x+60^0+k360^0\\3x=-x-60^0+k360^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=30^0+k180^0\\x=-15^0+k90^0\end{matrix}\right.\)
\(\Leftrightarrow sin\left(2x-90^0\right)=cos2x\)
\(\Leftrightarrow-cos2x=cos2x\)
\(\Rightarrow cos2x=0\Rightarrow2x=90^0+k180^0\)
\(\Rightarrow x=45^0+k90^0\)
\(cos^2x+sin^2x+2sinx.cosx=1+cos4x\)
\(\Leftrightarrow1+sin2x=1+cos4x\)
\(\Leftrightarrow cos4x=sin2x=cos\left(\frac{\pi}{2}-2x\right)\)
\(\Rightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}-2x+k2\pi\\4x=2x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{k\pi}{3}\\x=-\frac{\pi}{4}+k\pi\end{matrix}\right.\)
\(VT=\left(cos2x-cos4x\right)^2=\left(-2cos^22x+cos2x+1\right)^2\le\left(\dfrac{9}{8}\right)^2\\ VP=6+2sin3x\ge4\\ \rightarrow VT< VP\)
Suy ra phương trình vô nghiệm.