cos2 2x-\(\sqrt{3}sin4x=1+sin^22x\)
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TG
2
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 7 2021
b.
ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)
\(\sqrt{2}\left(sinx+cosx\right)=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}\)
\(\Leftrightarrow\sqrt{2}\left(sinx+cosx\right)=\dfrac{1}{sinx.cosx}\)
Đặt \(sinx+cosx=t\Rightarrow\left|t\right|\le\sqrt{2}\)
\(sinx.cosx=\dfrac{t^2-1}{2}\)
Pt trở thành:
\(\sqrt{2}t=\dfrac{2}{t^2-1}\Rightarrow t^3-t-\sqrt{2}=0\)
\(\Leftrightarrow\left(t-\sqrt{2}\right)\left(t^2+\sqrt{2}t+1\right)=0\)
\(\Leftrightarrow t=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow x+\dfrac{\pi}{4}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+k2\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 7 2021
a.
\(\Leftrightarrow sin^22x+cos^22x+\sqrt{3}sin4x+1+cos4x=0\)
\(\Leftrightarrow cos4x+\sqrt{3}sin4x=-2\)
\(\Leftrightarrow\dfrac{1}{2}cos4x+\dfrac{\sqrt{3}}{2}sin4x=-1\)
\(\Leftrightarrow cos\left(4x-\dfrac{\pi}{3}\right)=-1\)
\(\Leftrightarrow4x-\dfrac{\pi}{3}=\pi+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{3}+\dfrac{k\pi}{2}\)
28 tháng 6 2021
1.Pt \(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=sin\left(x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\left(\dfrac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\\2x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Rightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\)\(\left(k\in Z\right)\)
2.\(sin^22x+cos^23x=1\)
\(\Leftrightarrow\dfrac{1-cos4x}{2}+\dfrac{1+cos6x}{2}=1\)
\(\Leftrightarrow cos6x=cos4x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow x=\dfrac{k\pi}{5}\)\(\left(k\in Z\right)\) (Gộp nghiệm)
Vậy...
3. \(Pt\Leftrightarrow\left(sinx+sin3x\right)+\left(sin2x+sin4x\right)=0\)
\(\Leftrightarrow2.sin2x.cosx+2.sin3x.cosx=0\)
\(\Leftrightarrow2cosx\left(sin2x+sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=-sin2x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\sin3x=sin\left(\pi+2x\right)\end{matrix}\right.\)(\(k\in Z\))
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\)(\(k\in Z\))\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\) (\(k\in Z\))
Vậy...
4. Pt\(\Leftrightarrow\dfrac{1-cos2x}{2}+\dfrac{1-cos4x}{2}=\dfrac{1-cos6x}{2}\)
\(\Leftrightarrow cos2x+cos4x=1+cos6x\)
\(\Leftrightarrow2cos3x.cosx=2cos^23x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\cosx=cos3x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=-k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy...
DT
1
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 9 2020
ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow\left(tanx+cotx\right)^2=\frac{4+sin4x}{sin^22x}+2\)
\(\Leftrightarrow\left(\frac{sin^2x+cos^2x}{\frac{1}{2}sin2x}\right)^2=\frac{4+sin4x+2sin^22x}{sin^22x}\)
\(\Leftrightarrow\frac{4}{sin^22x}=\frac{4+sin4x+2sin^22x}{sin^22x}\)
\(\Leftrightarrow2sin^22x+sin4x=0\)
\(\Leftrightarrow1-cos4x+sin4x=0\)
\(\Leftrightarrow\sqrt{2}cos\left(4x+\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow cos\left(4x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\4x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\left(l\right)\\x=-\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 6 2020
\(\frac{sin3x+sinx+sin4x}{cos4x+1+cosx+cos3x}=\frac{2sin2x.cosx+2sin2x.cos2x}{2cos^22x+2cos2x.cosx}=\frac{2sin2x\left(cosx+cos2x\right)}{2cos2x\left(cos2x+cosx\right)}=\frac{sin2x}{cos2x}=tan2x\)
\(\frac{sin^22x+2cos\left(2\pi+\pi+2x\right)-2}{-3+4cos2x+cos\left(\pi-4x\right)}=\frac{sin^22x-2cos2x-2}{-3+4cos2x-cos4x}=\frac{4sin^2x.cos^2x-2\left(2cos^2x-1\right)-2}{-3+4\left(1-2sin^2x\right)-\left(1-2sin^22x\right)}\)
\(=\frac{4cos^2x\left(sin^2x-1\right)}{-8sin^2x+2sin^22x}=\frac{2cos^2x.\left(-cos^2x\right)}{-4sin^2x+4sin^2x.cos^2x}=\frac{cos^4x}{2sin^2x\left(1-cos^2x\right)}\)
\(=\frac{cos^4x}{2sin^4x}=\frac{1}{2}cot^4x\)
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 1
Coi như tất cả các biểu thức cần tính đạo hàm đều xác định.
1.
\(y'=2sin\sqrt{4x+3}.\left(sin\sqrt{4x+3}\right)'=2sin\sqrt{4x+3}.cos\sqrt{4x+3}.\left(\sqrt{4x+3}\right)'\)
\(=sin\left(2\sqrt{4x+3}\right).\dfrac{4}{2\sqrt{4x+3}}=\dfrac{2sin\left(2\sqrt{4x+3}\right)}{\sqrt{4x+3}}\)
2.
\(y'=3x^3+\dfrac{17}{x\sqrt{x}}\)
3.
\(y'=\dfrac{1}{2\sqrt{\dfrac{sin4x}{cos\left(x^2+2\right)}}}.\left(\dfrac{sin4x}{cos\left(x^2+2\right)}\right)'\)
\(=\dfrac{1}{2\sqrt{\dfrac{sin4x}{cos\left(x^2+2\right)}}}.\dfrac{4cos4x.cos\left(x^2+2\right)+2x.sin4x.sin\left(x^2+2\right)}{cos^2\left(x^2+2\right)}\)
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 1
4.
\(y'=-\dfrac{\left(\sqrt{sin^2\left(6-x\right)+4x}\right)'}{sin^2\left(6-x\right)+4x}=-\dfrac{\left[sin^2\left(6-x\right)+4x\right]'}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)
\(=-\dfrac{2sin\left(6-x\right).\left[sin\left(6-x\right)\right]'+4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}=-\dfrac{-2sin\left(6-x\right).cos\left(6-x\right)+4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)
\(=\dfrac{sin\left(12-2x\right)-4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)
5.
\(y'=sin^2\left(\dfrac{2x-1}{4-x}\right)+2x.sin\left(\dfrac{2x-1}{4-x}\right).\left[sin\left(\dfrac{2x-1}{4-x}\right)\right]'\)
\(=sin^2\left(\dfrac{2x-1}{4-x}\right)+2x.sin\left(\dfrac{2x-1}{4-x}\right).cos\left(\dfrac{2x-1}{4-x}\right).\left(\dfrac{2x-1}{4-x}\right)'\)
\(=sin^2\left(\dfrac{2x-1}{4-x}\right)+x.sin\left(\dfrac{4x-2}{4-x}\right).\dfrac{7}{\left(4-x\right)^2}\)
\(\Leftrightarrow\cos^22x-\sin^22x-\sqrt{3}\sin4x-1=0\)
\(\Leftrightarrow\cos^22x-\left(1-\cos^22x\right)-2\sqrt{3}\sin2x\cos2x-1=0\)
\(\Leftrightarrow2\cos^22x-2\sqrt{3}sin2x\cos2x-2=0\)
\(\Leftrightarrow\cos^22x-\sqrt{3}sin2x\cos2x=1\)
\(\Leftrightarrow\cos2x\left(\cos2x-\sqrt{3}sin2x\right)=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\cos2x=1\\\cos2x-\sqrt{3}\sin2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{k\sqcap}{2}\\\dfrac{1}{2}\cos2x-\dfrac{\sqrt{3}}{2}\sin2x=\dfrac{1}{2}\left(1\right)\end{matrix}\right.\)
(1) \(\Leftrightarrow\sin\dfrac{\sqcap}{6}\cos2x-\cos\dfrac{\sqcap}{6}\sin2x=\dfrac{1}{2}\)
\(\Leftrightarrow\sin\left(\dfrac{\sqcap}{6}-2x\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\sqcap}{6}-2x=\dfrac{\sqcap}{6}+k2\sqcap\\\dfrac{\sqcap}{6}-2x=\dfrac{5\sqcap}{6}+k2\sqcap\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\sqcap\\x=\dfrac{-\sqcap}{3}+k\sqcap\end{matrix}\right.\)
\(\Rightarrow S=\left\{{}\begin{matrix}\left[{}\begin{matrix}x=k\sqcap\\x=\dfrac{-\sqcap}{3}+k\sqcap\end{matrix}\right.\\x=\dfrac{k\sqcap}{2}\end{matrix}\right.\)