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NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 9 2020
e/
Đề câu này chắc chắn đúng chứ bạn?
f/
\(sin^4x+cos^4x=\frac{3}{4}\)
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\frac{3}{4}\)
\(\Leftrightarrow1-\frac{1}{2}\left(2sinx.cosx\right)^2=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}-\frac{1}{2}sin^22x=0\)
\(\Leftrightarrow1-2sin^22x=0\)
\(\Leftrightarrow cos4x=0\)
\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 9 2020
c/
\(y=sin\left(4x-\frac{\pi}{3}\right)+sin\left(\frac{\pi}{3}\right)+5\)
\(=sin\left(4x-\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}+5\)
Do \(-1\le sin\left(4x-\frac{\pi}{3}\right)\le1\)
\(\Rightarrow4+\frac{\sqrt{3}}{2}\le y\le6+\frac{\sqrt{3}}{2}\)
d/
\(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+3sin2x+5\)
\(y=6-3sin^2x.cos^2x+3sin2x\)
\(y=-\frac{3}{4}sin^22x+3sin2x+6\)
\(y=\frac{3}{4}\left(sin2x+1\right)\left(5-sin2x\right)+\frac{9}{4}\ge\frac{9}{4}\)
\(y_{min}=\frac{9}{4}\) khi \(sin2x=-1\)
\(y=\frac{3}{4}\left(sin2x-1\right)\left(3-sin2x\right)+\frac{33}{4}\le\frac{33}{4}\)
\(y_{max}=\frac{33}{4}\) khi \(sin2x=1\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 7 2020
a/
\(\Leftrightarrow sin2x\left(1+\sqrt{2}sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\1+\sqrt{2}sinx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sinx=-\frac{\sqrt{2}}{2}=sin\left(-\frac{\pi}{4}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\x=-\frac{\pi}{4}+k2\pi\\x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=-\frac{\pi}{4}+k2\pi\\x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow2sin2x.cos2x-\frac{1}{2}sin4x+\frac{1}{2}sinx=0\)
\(\Leftrightarrow sin4x-\frac{1}{2}sin4x+\frac{1}{2}sinx=0\)
\(\Leftrightarrow sin4x=-sinx=sin\left(-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}4x=-x+k2\pi\\4x=\pi+x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{5}\\x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 7 2020
e/
\(sin\left(\frac{3\pi}{2}-sinx\right)=1\)
\(\Leftrightarrow\frac{3\pi}{2}-sinx=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow sinx=\pi+k2\pi\)
Mà \(-1\le sinx\le1\Rightarrow-1\le\pi+k2\pi\le1\)
\(\Rightarrow\) Không tồn tại k nguyên thỏa mãn
Pt đã cho vô nghiệm
f/
\(cos^2x-sin^2x+sin4x=0\)
\(\Leftrightarrow cos2x+2sin2x.cos2x=0\)
\(\Leftrightarrow cos2x\left(1+2sin2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin2x=-\frac{1}{2}=sin\left(-\frac{\pi}{6}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\2x=-\frac{\pi}{6}+k2\pi\\2x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 2 2020
a/ Hmm, bạn có nhầm lẫn chỗ nào ko nhỉ, nghiệm của pt này xấu khủng khiếp
b/ \(\Leftrightarrow sin\frac{5x}{2}-cos\frac{5x}{2}-sin\frac{x}{2}-cos\frac{x}{2}=cos\frac{3x}{2}\)
\(\Leftrightarrow2cos\frac{3x}{2}.sinx-2cos\frac{3x}{2}cosx=cos\frac{3x}{2}\)
\(\Leftrightarrow cos\frac{3x}{2}\left(2sinx-2cosx-1\right)=0\)
\(\Leftrightarrow cos\frac{3x}{2}\left(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)-1\right)=0\)
c/ Do \(cosx\ne0\), chia 2 vế cho cosx ta được:
\(3\sqrt{tanx+1}\left(tanx+2\right)=5\left(tanx+3\right)\)
Đặt \(\sqrt{tanx+1}=t\ge0\)
\(\Leftrightarrow3t\left(t^2+1\right)=5\left(t^2+2\right)\)
\(\Leftrightarrow3t^3-5t^2+3t-10=0\)
\(\Leftrightarrow\left(t-2\right)\left(3t^2+t+5\right)=0\)
d/ \(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{3}\right)=-sin\left(2x-\frac{\pi}{3}\right)\)
Đặt \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
B
22 tháng 9 2020
a. \(sin\left(4x+\pi\right)=sin35^o\)
\(\Leftrightarrow sin\left(4x+180^o\right)=sin35^o\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+180^o=35^o+k.360^o,k\in Z\\4x+180^o=180^o-35^o+k.360^o,k\in Z\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=-145^o+k.360^o,k\in Z\\4x=-35^o+k.360^o,k\in Z\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{145^o}{4}+k.90,k\in Z\\x=-\frac{35^o}{4}+k.90^o,k\in Z\end{matrix}\right.\)
Vậy.....
b.\(sin4x=\frac{1}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=arcsin\left(\frac{1}{5}\right)+k2\pi,k\in Z\\4x=\pi-arcsin\left(\frac{1}{5}\right)+k2\pi,k\in Z\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{arcsin\left(\frac{1}{5}\right)}{4}+\frac{k\pi}{2},k\in Z\\x=\frac{\pi}{4}-\frac{arcsin\left(\frac{1}{5}\right)}{4}+\frac{k\pi}{2},k\in Z\end{matrix}\right.\)
Vậy....
B
22 tháng 9 2020
c. \(sin\left(x+\frac{8\pi}{7}\right)=3\)
Ta có: \(-1\le sinx\le1\)
\(\Rightarrow-1\le sin\left(3x+\frac{8\pi}{7}\right)\le1\)
Do đó phương trình trên vô nghiệm
d. \(sinx=-7\)
Ta có: \(-1\le sinx\le1\)
Do đó phương trình trên vô nghiệm
e. \(sin\left(3x+\pi\right)=sin\left(2x-3\pi\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\pi=2x-3\pi+k2\pi,k\in Z\\3x+\pi=\pi-2x+3\pi+k2\pi,k\in Z\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\pi+k2\pi,k\in Z\\5x=3\pi+k2\pi,k\in Z\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\pi+k2\pi,k\in Z\\x=\frac{3}{5}\pi+\frac{k2\pi}{5},k\in Z\end{matrix}\right.\)
Vậy......
f. \(sin\left(4x-\frac{\pi}{2}\right)=sin\left(\pi-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{2}=\pi-2x+k2\pi,k\in Z\\4x-\frac{\pi}{2}=\pi-\pi+2x+k2\pi,k\in Z\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}6x=\frac{3}{2}\pi+k2\pi,k\in Z\\2x=\frac{\pi}{2}+k2\pi,k\in Z\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{3},k\in Z\\x=\frac{\pi}{4}+k\pi,k\in Z\end{matrix}\right.\)
Vậy......
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2020
1.
ĐKXĐ: \(sin\left(2x+3\right)\ne0\Leftrightarrow2x+3\ne k\pi\)
\(\Leftrightarrow x\ne-\frac{3}{2}+\frac{k\pi}{2}\)
2.
ĐKXĐ: \(\left\{{}\begin{matrix}cos2x\ne0\\sinx\ne-1\\sin\left(3x+\frac{\pi}{6}\right)\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x\ne\frac{\pi}{2}+k\pi\\x\ne-\frac{\pi}{2}+k2\pi\\3x+\frac{\pi}{6}\ne k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{4}+\frac{k\pi}{2}\\x\ne-\frac{\pi}{2}+k2\pi\\x\ne-\frac{\pi}{18}+\frac{k\pi}{3}\end{matrix}\right.\)
3.
\(\left\{{}\begin{matrix}cos5x\ne0\\sin4x\ne cos3x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x\ne\frac{\pi}{2}+k\pi\\sin4x\ne sin\left(\frac{\pi}{2}-3x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{10}+\frac{k\pi}{5}\\4x\ne\frac{\pi}{2}-3x+k2\pi\\4x\ne\frac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{10}+\frac{k\pi}{5}\\x\ne\frac{\pi}{14}+\frac{k2\pi}{7}\\x\ne\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
NT
1
ta có:( theo công thức lượng giác nhân ba)
VT= cos3x.sin3x+sin3x.cos3x=cos3x.\(\frac{3sinx-sin3x}{4}\)+sin3x\(\frac{3cosx+cos3x}{4}\)
=\(\frac{3}{4}\)((sinx-\(\frac{1}{3}\)sin3x).cos3x+sin3x(cosx+\(\frac{1}{3}\)cos3x))
.=\(\frac{3}{4}\)(cos3x.sinx-\(\frac{1}{3}\)sin3x.cos3x+sin3x.cosx+\(\frac{1}{3}\)sin3x.cos3x)
=\(\frac{3}{4}\)(sinx.cos3x+cosx.sin3x)
=\(\frac{3}{4}\)sin(x+3x)=\(\frac{3}{4}\)sin4x
=> đpcm