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L
1
18 tháng 7 2020
Ta có:
\(-1\le\sin2x\le1\)
=> \(\sqrt{4-2.\left(1\right)^5}-8\le\sqrt{4-2.\left(\sin2x\right)^5}-8\le\sqrt{4-2.\left(-1\right)^5}-8\)
=> \(\sqrt{2}-8\le\sqrt{4-2.\left(\sin2x\right)^5}-8\le\sqrt{6}-8\)
=> tìm ddc min và max
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2020
c.
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos\left(8x+\frac{2\pi}{3}\right)=\frac{1}{2}-\frac{1}{2}cos\left(\frac{14\pi}{5}-2x\right)\)
\(\Leftrightarrow cos\left(8x+\frac{2\pi}{3}\right)=cos\left(2\pi+\frac{4\pi}{5}-2x\right)\)
\(\Leftrightarrow cos\left(8x+\frac{2\pi}{3}\right)=cos\left(\frac{4\pi}{5}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}8x+\frac{2\pi}{3}=\frac{4\pi}{5}-2x+k2\pi\\8x+\frac{2\pi}{3}=2x-\frac{4\pi}{5}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{75}+\frac{k\pi}{5}\\x=-\frac{11\pi}{45}+\frac{k\pi}{3}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2020
a.
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos4x=\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{2\pi}{3}\right)\)
\(\Leftrightarrow cos4x=-cos\left(2x+\frac{2\pi}{3}\right)\)
\(\Leftrightarrow cos4x=cos\left(\frac{\pi}{3}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{3}-2x+k2\pi\\4x=2x-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{18}+\frac{k\pi}{3}\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
b.
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos\left(10x+\frac{2\pi}{3}\right)-\frac{1}{2}-\frac{1}{2}cos\left(6x+\frac{\pi}{2}\right)=0\)
\(\Leftrightarrow cos\left(10x+\frac{2\pi}{3}\right)=-cos\left(6x+\frac{\pi}{2}\right)\)
\(\Leftrightarrow cos\left(10x+\frac{2\pi}{3}\right)=cos\left(\frac{\pi}{2}-6x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}10x+\frac{2\pi}{3}=\frac{\pi}{2}-6x+k2\pi\\10x+\frac{2\pi}{3}=6x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{96}+\frac{k\pi}{8}\\x=-\frac{7\pi}{24}+\frac{k\pi}{2}\end{matrix}\right.\)
5 tháng 10 2020
:v bn ns v là bn bik hết là dạng gì rr mà lm ko đc á :))
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 8 2020
d/
\(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}cos\left(\frac{x}{5}-\frac{\pi}{12}\right)-\frac{\sqrt{3}}{2}sin\left(\frac{x}{5}-\frac{\pi}{12}\right)\right)=sin\left(\frac{x}{5}+\frac{2\pi}{3}\right)-sin\left(\frac{3x}{5}+\frac{\pi}{6}\right)\)
\(\Leftrightarrow\sqrt{2}cos\left(\frac{x}{5}-\frac{\pi}{12}+\frac{\pi}{3}\right)=2cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)sin\left(\frac{\pi}{4}-\frac{x}{5}\right)\)
\(\Leftrightarrow cos\left(\frac{x}{5}-\frac{\pi}{4}\right)=\sqrt{2}cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)cos\left(\frac{x}{5}-\frac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\frac{x}{5}-\frac{\pi}{4}\right)=0\\cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{5}-\frac{\pi}{4}=\frac{\pi}{2}+k\pi\\\frac{2x}{5}+\frac{5\pi}{12}=\frac{\pi}{4}+k2\pi\\\frac{2x}{5}+\frac{5\pi}{12}=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{15\pi}{4}+k5\pi\\x=-\frac{5\pi}{12}+k5\pi\\x=-\frac{5\pi}{3}+k5\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 8 2020
c/
\(\Leftrightarrow\sqrt{3}sin\left(x-\frac{\pi}{3}\right)+cos\left(\frac{\pi}{3}-x\right)=2sin1972x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin\left(x-\frac{\pi}{3}\right)+\frac{1}{2}cos\left(x-\frac{\pi}{3}\right)=sin1972x\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}+\frac{\pi}{6}\right)=sin1972x\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=sin1972x\)
\(\Leftrightarrow\left[{}\begin{matrix}1972x=x-\frac{\pi}{6}+k2\pi\\1972x=\frac{7\pi}{6}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{11826}+\frac{k2\pi}{1971}\\x=\frac{7\pi}{11838}+\frac{k2\pi}{1973}\end{matrix}\right.\)
12 tháng 9 2021
1, \(y=2-sin\left(\dfrac{3x}{2}+x\right).cos\left(x+\dfrac{\pi}{2}\right)\)
\(y=2-\left(-cosx\right).\left(-sinx\right)\)
y = 2 - sinx.cosx
y = \(2-\dfrac{1}{2}sin2x\)
Max = 2 + \(\dfrac{1}{2}\) = 2,5
Min = \(2-\dfrac{1}{2}\) = 1,5
2, y = \(\sqrt{5-\dfrac{1}{2}sin^22x}\)
Min = \(\sqrt{5-\dfrac{1}{2}}=\dfrac{3\sqrt{2}}{2}\)
Max = \(\sqrt{5}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 9 2020
Câu g đề thiếu
Câu 2:
\(sin\left(2x+\frac{\pi}{6}\right)=\frac{2}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=arcsin\left(\frac{2}{5}\right)+k2\pi\\2x+\frac{\pi}{6}=\pi-arcsin\left(\frac{2}{5}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+\frac{1}{2}arcsin\left(\frac{2}{5}\right)+k\pi\\x=\frac{5\pi}{12}-\frac{1}{2}arcsin\left(\frac{2}{5}\right)+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\approx-0,056\left(rad\right)\\x\approx1,1\left(rad\right)\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 10 2020
a.
\(cos\left(3x-\frac{\pi}{6}\right)=sin\left(2x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(3x-\frac{\pi}{6}\right)=cos\left(\frac{\pi}{6}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{6}=\frac{\pi}{6}-2x+k2\pi\\3x-\frac{\pi}{6}=2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\cos3x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\cos2x\ne\frac{1}{2}\end{matrix}\right.\)
\(tan3x-tanx=0\)
\(\Leftrightarrow\frac{sin3x}{cos3x}-\frac{sinx}{cosx}=0\)
\(\Leftrightarrow sin3x.cosx-cos3x.sinx=0\)
\(\Leftrightarrow sin2x=0\)
\(\Leftrightarrow2sinx.cosx=0\)
\(\Leftrightarrow sinx=0\Leftrightarrow x=k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 10 2020
c.
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{2\pi}{5}\right)=\frac{1}{2}-\frac{1}{2}cos\left(4x+\frac{8\pi}{5}\right)\)
\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=-cos\left(4x+\frac{3\pi}{5}+\pi\right)\)
\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=cos\left(4x+\frac{3\pi}{5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{3\pi}{5}=2x-\frac{2\pi}{5}+k2\pi\\4x+\frac{3\pi}{5}=\frac{2\pi}{5}-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
d.
\(\Leftrightarrow cos^2\left(2x-1\right)=0\)
\(\Leftrightarrow cos\left(2x-1\right)=0\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{1}{2}+\frac{k\pi}{2}\)