Sin(x-π/4)= cos3x
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Giải phương trình
1 : sin2x = cos3x
2 : cos(2x - \(\frac{\text{π}}{4}\) ) + sin(x+ \(\frac{\text{π}}{4}\)) = 0
Giảt pt 1,sin(4x-10°) = √2/2 2, cos(2x=7/8 3, tan 2x=tanx 4, cot(x+π/5)=-1 5, cos3x=sin5x
1.
\(sin\left(4x-10^0\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(4x-10^0\right)=sin45^0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-10^0=45^0+k360^0\\4x-10^0=135^0+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=55^0+k360^0\\4x=145^0+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=13,75^0+k90^0\\x=36,25^0+k90^0\end{matrix}\right.\) (\(k\in Z\))
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2.
Đề không đúng
3.
ĐKXĐ: \(\left\{{}\begin{matrix}cos2x\ne0\\cosx\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(tan2x=tanx\)
\(\Rightarrow2x=x+k\pi\)
\(\Rightarrow x=k\pi\)
4.
\(cot\left(x+\dfrac{\pi}{5}\right)=-1\)
\(\Leftrightarrow x+\dfrac{\pi}{5}=-\dfrac{\pi}{4}+k\pi\)
\(\Leftrightarrow x=-\dfrac{9\pi}{20}+k\pi\) (\(k\in Z\))
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5.
\(cos3x=sin5x\)
\(\Leftrightarrow sin5x=sin\left(\dfrac{\pi}{2}-3x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{2}-3x+k2\pi\\5x=\dfrac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}8x=\dfrac{\pi}{2}+k2\pi\\2x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{4}\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\) (\(k\in Z\))
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sin (2x+π/3) +cos3x = 0
\(sin\left(2x+\dfrac{\pi}{3}\right)+cos3x=0\)
\(\Leftrightarrow cos\left(\dfrac{\pi}{6}-2x\right)+cos3x=0\)
\(\Leftrightarrow2cos\left(\dfrac{\pi}{12}+\dfrac{x}{2}\right).cos\left(\dfrac{\pi}{12}-\dfrac{5x}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\dfrac{\pi}{12}+\dfrac{x}{2}\right)=0\\cos\left(\dfrac{\pi}{12}-\dfrac{5x}{2}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{12}+\dfrac{x}{2}=\dfrac{\pi}{2}+k\pi\\\dfrac{\pi}{12}-\dfrac{5x}{2}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{6}+k2\pi\\x=-\dfrac{\pi}{6}+\dfrac{k2\pi}{5}\end{matrix}\right.\)
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Ta có: \(sin\left(2x+\dfrac{\pi}{3}\right)=-cos3x=cos\left(\pi-3x\right)=sin\left(\dfrac{\pi}{2}-\left(\pi-3x\right)\right)=sin\left(3x-\dfrac{1}{2}\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=3x-\dfrac{1}{2}+k2\pi\\2x+\dfrac{\pi}{3}=\pi-3x+\dfrac{1}{2}+k2\pi\end{matrix}\right.\) Bạn tự tìm x được.
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Tính:F=Cos(π/4+α) x cos(π/4-α)
G=Sin(π/3+α) x cos(π/3-α)
H=cos(π/2-α) x sin(π/2+α)
I=sin(π/4+α) - cos(π/4-α)
K=cos(π/6-x) - sin(π/3+x)
giải các phương trình sau :1. sin( x+pi/4)2/32.cos2x-5sinx-303.cos3xsin2x4.cos3x-sqrt{ }3 với -pi/2x05.4sin^4x + 12cos^2x76.cot(x-1)(cos2x)/(1+tanx) + sin^2x - 1/2sin2x7.sin^23x-cos^24xsin^25x-cos^26x
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giải các phương trình sau :
1. sin( x+\(\pi\)/4)=2/3
2.cos2x-5sinx-3=0
3.cos3x=sin2x
4.cos3x=-\(\sqrt{ }\)3 với -\(\pi\)/2<x<0
5.4sin\(^4\)x + 12cos\(^2\)x=7
6.cot(x-1)=(cos2x)/(1+tanx) + sin\(^2\)x - 1/2sin2x
7.sin\(^2\)3x-cos\(^2\)4x=sin\(^2\)5x-cos\(^2\)6x
sin ( x + π 4 ) + sin ( x − π 4 ) = 0 thuộc khoảng (0;4π)
\(cos3x=sin\left(x+\dfrac{\sqcap}{4}\right)\)
\(cos3x=sin\left(x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow cos3x=cos\left(\dfrac{\pi}{4}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{4}-x+k2\pi\\3x=-\dfrac{\pi}{4}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{8}+k\pi\end{matrix}\right.\)
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Pt\(\Rightarrow cos3x=cos[\dfrac{\pi}{2}-(x+\dfrac{\pi}{4})]\)
\(\Rightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{2}-(x+\dfrac{\pi}{4})+k2\pi\\3x=-\dfrac{\pi}{2}+\left(x+\dfrac{\pi}{4}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{4}+k2\pi\\2x=-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+k2\pi\\x=-\dfrac{\pi}{8}+k2\pi\end{matrix}\right.\)(k\(\in\)Z)
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giai Pt sin(x+π/4) + sin(2x+π/4)=2sinv/4
Giai các pt sau
1. \(\sqrt{3}\cos5x-2\sin3x.\cos2x-\sin x=0\)
4. \(\sin3x+\cos3x-\sin x+\cos x=\sqrt{2}\cos2x\)
6. \(\sin x+\cos x.\sin2x+\sqrt{3}\cos3x=2\left(\cos4x+\sin x^3\right)\)