\(\frac{1-sin^2cos^2x}{cos^2x}-cos^2x\) \(\sqrt{sin^4x+4cos^2x}+\sqrt{cos^4+4sin^2x}\)
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NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 9 2020
e/
\(2cos^2x+2cos^22x+4cos^32x-3cos2x=5\)
\(\Leftrightarrow1+cos2x+2cos^22x+4cos^32x-3cos2x=5\)
\(\Leftrightarrow2cos^32x+cos^22x-cos2x-2=0\)
\(\Leftrightarrow\left(cos2x-1\right)\left(2cos^22x+3cos2x+2\right)=0\)
\(\Leftrightarrow cos2x=1\)
\(\Leftrightarrow x=k\pi\)
7 tháng 12 2023
\(0< =sin^2x< =1\)
=>\(-2< =sin^2x-2< =-1\)
=>\(sin^2x-2< 0\)
\(0< =cos^2x< =1\)
=>\(-2< =cos^2x-2< =-1\)
\(\Leftrightarrow cos^2x-2< 0\)
\(\sqrt{sin^4x+4cos^2x}+\sqrt{cos^4x+4\cdot sin^2x}\)
\(=\sqrt{sin^4x+4\left(1-sin^2x\right)}+\sqrt{cos^4x+4\cdot\left(1-cos^2x\right)}\)
\(=\sqrt{sin^4x-4sin^xx+4}+\sqrt{cos^4x-4\cdot cos^2x+4}\)
\(=\sqrt{\left(sin^2x-2\right)^2}+\sqrt{\left(cos^2x-2\right)^2}\)
\(=\left|sin^2x-2\right|+\left|cos^2x-2\right|\)
\(=2-sin^2x+2-cos^2x\)
\(=4-\left(sin^2x+cos^2x\right)=4-1=3\)
NC
1
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 6 2020
\(=\sqrt{sin^4x+4\left(1-sin^2x\right)}+\sqrt{cos^4x+4\left(1-cos^2x\right)}\)
\(=\sqrt{4-4sin^2x+sin^4x}+\sqrt{4-4cos^2x+cos^4x}\)
\(=\sqrt{\left(2-sin^2x\right)^2}+\sqrt{\left(2-cos^2x\right)^2}\)
\(=2-sin^2x+2-cos^2x=4-\left(sin^2x+cos^2x\right)\)
\(=3\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 8 2020
d/
\(2cos^22x+cos2x=4sin^22x.cos^2x\)
\(\Leftrightarrow2cos^22x+cos2x=2\left(1+cos2x\right)\left(1-cos^22x\right)\)
\(\Leftrightarrow2cos^32x+4cos^22x-cos2x-2=0\)
\(\Leftrightarrow\left(cos2x+2\right)\left(2cos^22x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-2\left(vn\right)\\2cos^22x-1=0\end{matrix}\right.\)
\(\Leftrightarrow cos4x=0\)
\(\Leftrightarrow4x=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 8 2020
c/
\(cos^4x+sin^6x=cos2x\)
\(\Leftrightarrow\left(\frac{1+cos2x}{2}\right)^2+\left(\frac{1-cos2x}{2}\right)^3=cos2x\)
\(\Leftrightarrow cos^32x-5cos^2x+7cos2x-3=0\)
\(\Leftrightarrow\left(cos2x-1\right)^2\left(cos2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=1\\cos2x=3\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow2x=k2\pi\)
\(\Rightarrow x=k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 4 2019
\(\sqrt{sin^4x+4\left(1-sin^2x\right)}+\sqrt{cos^4x+4\left(1-cos^2x\right)}\)
\(=\sqrt{sin^4x-4sin^2x+4}+\sqrt{cos^4x-4cos^2x+4}\)
\(=\sqrt{\left(2-sin^2x\right)^2}+\sqrt{\left(2-cos^2x\right)^2}\)
\(=2-sin^2x+2-cos^2x\)
\(=4-\left(sin^2x+cos^2x\right)=3\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 7 2021
1a.
Đặt \(5x+6=u\)
\(cos2u+4\sqrt{2}sinu-4=0\)
\(\Leftrightarrow1-2sin^2u+4\sqrt{2}sinu-4=0\)
\(\Leftrightarrow2sin^2u-4\sqrt{2}sinu+3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinu=\dfrac{3\sqrt{2}}{2}>1\left(loại\right)\\sinu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow sin\left(5x+6\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+6=\dfrac{\pi}{4}+k2\pi\\5x+6=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}+\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=-\dfrac{6}{5}+\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 7 2021
1b.
Đặt \(2x+1=u\)
\(cos2u+3sinu=2\)
\(\Leftrightarrow1-2sin^2u+3sinu=2\)
\(\Leftrightarrow2sin^2u-3sinu+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinu=1\\sinu=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(2x+1\right)=1\\sin\left(2x+1\right)=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{\pi}{2}+k2\pi\\2x+1=\dfrac{\pi}{6}+k2\pi\\2x+1=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}+\dfrac{\pi}{4}+k\pi\\x=-\dfrac{1}{2}+\dfrac{\pi}{12}+k\pi\\x=-\dfrac{1}{2}+\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 7 2020
e/
\(y=5sinx+6cosx-7\)
\(=\sqrt{61}\left(\frac{5}{\sqrt{61}}sinx+\frac{6}{\sqrt{61}}cosx\right)-7\)
\(=\sqrt{61}\left(sinx.cosa+cosx.sina\right)-7\) (với \(a\in\left(0;\pi\right)\) sao cho \(cosa=\frac{5}{\sqrt{61}}\))
\(=\sqrt{61}.sin\left(x+a\right)-7\)
Do \(-1\le sin\left(x+a\right)\le1\Rightarrow7-\sqrt{61}\le y\le7+\sqrt{61}\)
\(y_{min}=7-\sqrt{61}\) khi \(sin\left(x+a\right)=-1\)
\(y_{max}=7+\sqrt{61}\) khi \(sin\left(x+a\right)=1\)
f/
\(y=2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)+3\)
\(=2sin\left(x+\frac{\pi}{3}\right)+3\)
\(\Rightarrow1\le y\le5\)
\(y_{min}=1\) khi \(sin\left(x+\frac{\pi}{3}\right)=-1\)
\(y_{max}=5\) khi \(x+\frac{\pi}{3}=1\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 7 2020
c/
\(y=2\left(1-cos2x\right)+sin2x+cos2x\)
\(=sin2x-cos2x+2=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)+2\)
Do \(-1\le sin\left(2x-\frac{\pi}{4}\right)\le1\)
\(\Rightarrow2-\sqrt{2}\le y\le2+\sqrt{2}\)
\(y_{min}=2-\sqrt{2}\) khi \(sin\left(2x-\frac{\pi}{4}\right)=-1\)
\(y_{max}=2+\sqrt{2}\) khi \(sin\left(2x+\frac{\pi}{4}\right)=1\)
d/
\(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)\)
\(=1-3sin^2x.cos^2x\)
\(=1-\frac{3}{4}sin^22x\)
Mà \(0\le sin^22x\le1\Rightarrow\frac{1}{4}\le y\le1\)
\(y_{min}=\frac{1}{4}\) khi \(sin^22x=1\)
\(y_{max}=1\) khi \(sin2x=0\)