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Ryoji
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Nguyễn Việt Lâm
7 tháng 5 2019 lúc 15:46

\(\frac{sin^22x+4sin^2x-4}{1-8sin^2x-cos4x}=\frac{4sin^2x.cos^2x-4\left(1-sin^2x\right)}{1-8sin^2x-\left(1-2sin^22x\right)}=\frac{4sin^2x.cos^2x-4cos^2x}{2sin^22x-8sin^2x}\)

\(=\frac{-4cos^2x\left(1-sin^2x\right)}{8sin^2x.cos^2x-8sin^2x}=\frac{-4cos^2x.cos^2x}{-8sin^2x\left(1-cos^2x\right)}=\frac{cos^4x}{2sin^4x}=\frac{1}{2}cot^4x\)

\(\frac{cos2x}{cot^2x-tan^2x}=\frac{cos2x.sin^2x.cos^2x}{cos^4x-sin^4x}=\frac{\left(cos^2x-sin^2x\right).\left(2sinx.cosx\right)^2}{4\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)}=\frac{1}{4}sin^22x\)

Tường Nguyễn Thế
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Nguyễn Việt Lâm
18 tháng 4 2019 lúc 23:36

\(A=\frac{1}{2}\left(\frac{sin^2x}{cos^2x}-1\right)\frac{cosx}{sinx}+cos4x.cot2x+sin4x\)

\(A=\frac{-1}{2}\left(\frac{cos^2x-sin^2x}{cos^2x}\right)\frac{cosx}{sinx}+cos4x.cot2x+sin4x\)

\(A=\frac{-cos2x}{2cosx.sinx}+cos4x.cot2x+sin4x\)

\(A=-cot2x+cos4x.cot2x+sin4x\)

\(A=cot2x\left(cos4x-1\right)+sin4x\)

\(A=\frac{cos2x}{sin2x}.\left(1-2sin^22x-1\right)+sin4x\)

\(A=\frac{-2cos2x.sin^22x}{sin2x}+sin4x\)

\(A=-sin4x+sin4x=0\)

Nguyễn Ngọc Trâm
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Nguyễn Việt Lâm
5 tháng 6 2020 lúc 14:22

\(\frac{1+sin4x+cos4x}{1-sin4x+cos4x}=\frac{1+2sin2x.cos2x+2cos^22x-1}{1-2sin2x.cos2x+2cos^22x-1}\)

\(=\frac{2cos2x\left(sin2x+cos2x\right)}{2cos2x\left(cos2x-sin2x\right)}=\frac{sin2x+cos2x}{cos2x-sin2x}\)

\(=\frac{\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)}{\sqrt{2}cos\left(2x+\frac{\pi}{4}\right)}=tan\left(2x+\frac{\pi}{4}\right)\)

\(\left(sin5x-cos5x\right)^2-\left(sin3x+cos3x\right)^2\)

\(=\left(\sqrt{2}sin\left(5x-\frac{\pi}{4}\right)\right)^2-\left(\sqrt{2}sin\left(3x+\frac{\pi}{4}\right)\right)^2\)

\(=2sin^2\left(5x-\frac{\pi}{4}\right)-2sin^2\left(3x+\frac{\pi}{4}\right)\)

\(=1-cos\left(10x-\frac{\pi}{2}\right)-1+cos\left(6x+\frac{\pi}{2}\right)\)

\(=-sin10x-sin6x=-2sin8x.cos2x\)

nguyễn hoàng lê thi
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Nguyễn Việt Lâm
2 tháng 8 2020 lúc 20:52

28.

\(tan\left(2x-15^0\right)=1\Leftrightarrow2x-15^0=45^0+k180^0\)

\(\Leftrightarrow2x=60^0+k180^0\)

\(\Leftrightarrow x=30^0+k90^0\)

\(-90^0\le30^0+k90^0\le90^0\Rightarrow k=\left\{-1;0\right\}\)

\(\Rightarrow x=\left\{-60^0;30^0\right\}\Rightarrow\sum x=-30^0\)

34.

\(tan\left(x+\frac{\pi}{2}\right)=1\Leftrightarrow x+\frac{\pi}{2}=\frac{\pi}{4}+k\pi\)

\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)

\(\Rightarrow sin\left(2x-\frac{\pi}{6}\right)=sin\left[2\left(-\frac{\pi}{4}+k\pi\right)-\frac{\pi}{6}\right]\)

\(=sin\left(-\frac{2\pi}{3}+k2\pi\right)=sin\left(-\frac{2\pi}{3}\right)=-\frac{\sqrt{3}}{2}\)

Bình Trần Thị
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Thanh Thuy
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Lê Phương Thảo
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Nguyễn Việt Lâm
4 tháng 10 2020 lúc 15:34

1.

\(\Leftrightarrow3x=k\pi\Leftrightarrow x=\frac{k\pi}{3}\)

2.

\(\Leftrightarrow cos5x=0\Leftrightarrow5x=\frac{\pi}{2}+k\pi\Leftrightarrow x=\frac{\pi}{10}+\frac{k\pi}{5}\)

4.

\(cos3x+cosx+cos2x=0\)

\(\Leftrightarrow2cos2x.cosx+cos2x=0\)

\(\Leftrightarrow cos2x\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cosx=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

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Nguyễn Việt Lâm
4 tháng 10 2020 lúc 15:35

3. ĐKXĐ: ...

\(\Leftrightarrow\frac{sin\left(x-15\right)}{cos\left(x-15\right)}=\frac{3sin\left(x+15\right)}{cos\left(x+15\right)}\)

\(\Leftrightarrow sin\left(x-15\right)cos\left(x+15\right)=3sin\left(x+15\right)cos\left(x-15\right)\)

\(\Leftrightarrow sin2x-sin30^0=3\left[sin2x+sin30^0\right]\)

\(\Leftrightarrow sin2x-\frac{1}{2}=3sin2x+\frac{3}{2}\)

\(\Leftrightarrow sin2x=-1\)

\(\Leftrightarrow2x=-\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=-\frac{\pi}{4}+k\pi\)

Nguyễn Việt Lâm
4 tháng 10 2020 lúc 15:38

5.

\(sin6x+sin2x+sin4x=0\)

\(\Leftrightarrow2sin4x.cos2x+sin4x=0\)

\(\Leftrightarrow sin4x\left(2cos2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\cos2x=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{4}\\x=\pm\frac{\pi}{3}+k\pi\end{matrix}\right.\)

6. ĐKXĐ; ...

\(\Leftrightarrow tanx+tan2x=1-tanx.tan2x\)

\(\Leftrightarrow\frac{tanx+tan2x}{1-tanx.tan2x}=1\)

\(\Leftrightarrow tan3x=1\)

\(\Leftrightarrow x=\frac{\pi}{12}+\frac{k\pi}{3}\)

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tran duc huy
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Nguyễn Việt Lâm
8 tháng 8 2020 lúc 7:48

1. ĐKXĐ: ...

\(\Leftrightarrow tan\left(x+\frac{\pi}{3}\right)=\frac{1}{tan\left(2x-\frac{\pi}{4}\right)}\)

\(\Leftrightarrow tan\left(x+\frac{\pi}{3}\right)=cot\left(2x-\frac{\pi}{4}\right)\)

\(\Leftrightarrow tan\left(x+\frac{\pi}{3}\right)=tan\left(\frac{3\pi}{4}-2x\right)\)

\(\Leftrightarrow x+\frac{\pi}{3}=\frac{3\pi}{4}-2x+k\pi\)

\(\Rightarrow x=\frac{5\pi}{36}+\frac{k\pi}{3}\)

2.

ĐKXĐ: ...

\(\Leftrightarrow tan\left(x+1\right)=\frac{1}{cot\left(2x+3\right)}\)

\(\Leftrightarrow tan\left(x+1\right)=tan\left(2x+3\right)\)

\(\Leftrightarrow2x+3=x+1+k\pi\)

\(\Rightarrow x=-2+k\pi\)

Nguyễn Việt Lâm
8 tháng 8 2020 lúc 7:52

3.

ĐKXĐ: ...

\(\Leftrightarrow tan^22x+\left(\frac{1}{cos^22x}+1\right)=8\)

\(\Leftrightarrow tan^22x+tan^22x=8\)

\(\Leftrightarrow tan^22x=4\)

\(\Rightarrow\left[{}\begin{matrix}tan2x=2\\tan2x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=arctan\left(2\right)+k180^0\\2x=-arctan\left(2\right)+k180^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}arctan\left(2\right)+k90^0\\x=-\frac{1}{2}arctan\left(2\right)+k90^0\end{matrix}\right.\)

Nghiệm trên nhận các giá trị \(k=\left\{0;1;2;3\right\}\) ; nghiệm dưới nhận các giá trị \(k=\left\{1;2;3;4\right\}\)

nguyễn thế minh
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