Giải pt sau:
\(sin\frac{x}{2}cos\frac{\pi}{4}-cos\frac{x}{3}sin\frac{\pi}{4}=0\)
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NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 7 2020
c/
\(\Leftrightarrow cos^3\left(x-\frac{\pi}{3}\right)=\frac{1}{8}\)
\(\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=cos\left(\frac{\pi}{3}\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{3}+k2\pi\\x-\frac{\pi}{3}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 7 2020
a/
\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{6}-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+k2\pi\\2x-\frac{\pi}{3}=x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{\pi}{6}+k2\pi\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{6}+\frac{k2\pi}{3}\)
b/
\(\Rightarrow sin^4x-cos^4x=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow-cos2x=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos2x=-sin\left(x+\frac{\pi}{3}\right)=cos\left(x+\frac{5\pi}{6}\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x=x+\frac{5\pi}{6}+k2\pi\\2x=-x-\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{5\pi}{6}+k2\pi\\x=-\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a)
\(\begin{array}{l}\sin \left( {2x - \frac{\pi }{6}} \right) = - \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{3}} \right)\end{array}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{6} = - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{6} = \pi + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = - \frac{\pi }{6} + k2\pi \\2x = \frac{{3\pi }}{2} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{{12}} + k\pi \\x = \frac{{3\pi }}{4} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
b) \(\begin{array}{l}\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\\ \Leftrightarrow \cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{3}\end{array}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{{3x}}{2} + \frac{\pi }{4} = \frac{\pi }{3} + k2\pi \\\frac{{3x}}{2} + \frac{\pi }{4} = \frac{{ - \pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{18}} + \frac{{k4\pi }}{3}\\x = \frac{{ - 7\pi }}{{18}} + \frac{{k4\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
c)
\(\begin{array}{l}\sin 3x - \cos 5x = 0\\ \Leftrightarrow \sin 3x = \cos 5x\\ \Leftrightarrow \cos 5x = \cos \left( {\frac{\pi }{2} - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} - 3x + k2\pi \\5x = - \left( {\frac{\pi }{2} - 3x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}8x = \frac{\pi }{2} + k2\pi \\2x = - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{16}} + \frac{{k\pi }}{4}\\x = - \frac{\pi }{4} + k\pi \end{array} \right.\end{array}\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
d)
\(\begin{array}{l}{\cos ^2}x = \frac{1}{4}\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{1}{2}\\\cos x = - \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \cos \frac{\pi }{3}\\\cos x = \cos \frac{{2\pi }}{3}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x = - \frac{\pi }{3} + k2\pi \end{array} \right.\\\left[ \begin{array}{l}x = \frac{{2\pi }}{3} + k2\pi \\x = - \frac{{2\pi }}{3} + k2\pi \end{array} \right.\end{array} \right.\end{array}\)
e)
\(\begin{array}{l}\sin x - \sqrt 3 \cos x = 0\\ \Leftrightarrow \frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{3}.\sin x - \sin \frac{\pi }{3}.\cos x = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = \sin 0\\ \Leftrightarrow x - \frac{\pi }{3} = k\pi ;k \in Z\\ \Leftrightarrow x = \frac{\pi }{3} + k\pi ;k \in Z\end{array}\)
f)
\(\begin{array}{l}\sin x + \cos x = 0\\ \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin x + \frac{{\sqrt 2 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{4}.\sin x + \sin \frac{\pi }{4}.\cos x = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin 0\\ \Leftrightarrow x + \frac{\pi }{4} = k\pi ;k \in Z\\ \Leftrightarrow x = - \frac{\pi }{4} + k\pi ;k \in Z\end{array}\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 9 2020
a.
\(\Leftrightarrow\left[{}\begin{matrix}3x=90^0-x+k360^0\\3x=90^0+x+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{45^0}{2}+k90^0\\x=45^0+k180^0\end{matrix}\right.\)
b.
\(\Leftrightarrow cos\left(3x+45^0\right)=cos\left(x-180^0\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+45^0=x-180^0+k360^0\\3x+45^0=180^0-x+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{225^0}{2}+k180^0\\x=\frac{135^0}{4}+k90^0\end{matrix}\right.\)
c.
\(\Leftrightarrow sin\left(2x+\frac{\pi}{3}\right)=sin\left(-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{3}=-x+k2\pi\\2x+\frac{\pi}{3}=\pi+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{9}+\frac{k2\pi}{3}\\x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 9 2020
d.
\(\Leftrightarrow sin\left(x-\frac{2\pi}{3}\right)=cos2x\)
\(\Leftrightarrow sin\left(x-\frac{2\pi}{3}\right)=sin\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{2\pi}{3}=\frac{\pi}{2}-x+k2\pi\\x-\frac{2\pi}{3}=2x+\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{12}+k\pi\\x=-\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
e.
\(\Leftrightarrow cos\left(2x-\frac{\pi}{4}\right)=sin\left(2x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(2x-\frac{\pi}{4}\right)=cos\left(\frac{\pi}{6}-2x\right)\)
\(\Leftrightarrow2x-\frac{\pi}{4}=\frac{\pi}{6}-2x+k2\pi\)
\(\Leftrightarrow x=\frac{5\pi}{48}+\frac{k\pi}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 4 2019
\(A=cosa\left(sinb.cosc-cosb.sinc\right)+cosb\left(sinc.cosa-cosc.sina\right)+cosc\left(sinacosb-cosasinb\right)\)
\(A=cosasinbcosc-cosacosbsinc+cosacosbsinc-sinacosbcosc+sinacosbcosc-cosasinbcosc\)
\(A=0\)
\(B=sin^2x+\frac{1}{2}\left(cos\frac{2\pi}{3}+cos2x\right)\)
\(B=\frac{1}{2}-\frac{1}{2}cos2x-\frac{1}{4}+\frac{1}{2}cos2x\)
\(B=\frac{1}{4}\)
\(C=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}+2x\right)+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}-2x\right)\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x-\frac{1}{2}\left(cos\left(\frac{4\pi}{3}+2x\right)+cos\left(\frac{4\pi}{3}-2x\right)\right)\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x-cos\frac{4\pi}{3}.cos2x\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x\)
\(C=\frac{3}{2}\)
\(D=\frac{1}{2}\left[\sqrt{2}sin\left(\frac{\pi}{4}+x\right)\right]^2-sin^2x-sinx.\sqrt{2}cos\left(\frac{\pi}{4}+x\right)\)
\(D=\frac{1}{2}\left(sinx+cosx\right)^2-sin^2x-sinx\left(sinx-cosx\right)\)
\(D=\frac{1}{2}\left(1+2sinx.cosx\right)-sin^2x-sin^2x+sinx.cosx\)
\(D=\frac{1}{2}+sinxcosx+sinxcosx=\frac{1}{2}+sin2x\)
30 tháng 4 2019
Góc độ cao của thang dựa vào tường là 60º và chân thang cách tường 4,6 m. Chiều dài của thang là
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 10 2020
Câu 2 bạn coi lại đề
3.
\(1+2sinx.cosx-2cosx+\sqrt{2}sinx+2cosx\left(1-cosx\right)=0\)
\(\Leftrightarrow sin2x-\left(2cos^2x-1\right)+\sqrt{2}sinx=0\)
\(\Leftrightarrow sin2x-cos2x=-\sqrt{2}sinx\)
\(\Leftrightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}sin\left(-x\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=sin\left(-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-x+k2\pi\\2x-\frac{\pi}{4}=\pi+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 10 2020
4.
Bạn coi lại đề, xuất hiện 2 số hạng \(cos4x\) ở vế trái nên chắc là bạn ghi nhầm
5.
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)-1\)
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=cos\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=sin2x\)
\(\Leftrightarrow sin2x\left(sinx-cosx.sin2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\Leftrightarrow x=...\\sinx-cosx.sin2x-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)
\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)
\(\Leftrightarrow2sin^3x-sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)
\(\Leftrightarrow...\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a)
\(\sin \left( {2x + \frac{\pi }{4}} \right) = \sin x \Leftrightarrow \left[ \begin{array}{l}2x + \frac{\pi }{4} = x + k2\pi \\2x + \frac{\pi }{4} = \pi - x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{4} + k2\pi \\3x = \pi - \frac{\pi }{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{4} + k2\pi \\x = \frac{\pi }{4} + \frac{{k2\pi }}{3}\end{array} \right.;k \in Z\)
b)
\(\begin{array}{l}\sin 2x = \cos 3x\\ \Leftrightarrow \cos 3x = \cos \left( {\frac{\pi }{2} - 2x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}3x = \frac{\pi }{2} - 2x + k2\pi \\3x = - \left( {\frac{\pi }{2} - 2x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} + k2\pi \\x = - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{10}} + \frac{{k2\pi }}{5}\\x = - \frac{\pi }{2} + k2\pi \end{array} \right.\end{array}\)
c)
\(\begin{array}{l}{\cos ^2}2x = {\cos ^2}\left( {x + \frac{\pi }{6}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right)\\\cos 2x = - \cos \left( {x + \frac{\pi }{6}} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right)\\\cos 2x = \cos \left( {\pi - \left( {x + \frac{\pi }{6}} \right)} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right)\\\cos 2x = \cos \left( {\frac{{5\pi }}{6} - x} \right)\end{array} \right.\end{array}\)
Với \(\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right) \Leftrightarrow \left[ \begin{array}{l}2x = - \left( {x + \frac{\pi }{6}} \right) + k2\pi \\2x = x + \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}3x = - \frac{\pi }{6} + k2\pi \\x = \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{{18}} + \frac{{k2\pi }}{3}\\x = \frac{\pi }{6} + k2\pi \end{array} \right.\)
Với \(\cos 2x = \cos \left( {\frac{{5\pi }}{6} - x} \right) \Leftrightarrow \left[ \begin{array}{l}2x = \frac{{5\pi }}{6} - x + k2\pi \\2x = - \left( {\frac{{5\pi }}{6} - x} \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}3x = \frac{{5\pi }}{6} + k2\pi \\x = - \frac{{5\pi }}{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{5\pi }}{{18}} + \frac{{k2\pi }}{3}\\x = - \frac{{5\pi }}{6} + k2\pi \end{array} \right.\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
\(\begin{array}{l}A = \cos \left( {x + \frac{\pi }{6}} \right)\cos \left( {x - \frac{\pi }{6}} \right) = \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{6} + x - \frac{\pi }{6}} \right) + \cos \left( {x + \frac{\pi }{6} - x + \frac{\pi }{6}} \right)} \right]\\A = \frac{1}{2}\left[ {\cos 2x + \cos \frac{\pi }{3}} \right] = \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) = \frac{3}{8}\end{array}\)
\(\begin{array}{l}B = \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right) = - \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{3} + x - \frac{\pi }{3}} \right) - \cos \left( {x + \frac{\pi }{3} - x + \frac{\pi }{3}} \right)} \right]\\B = - \frac{1}{2}\left( {\cos 2x - \cos \frac{{2\pi }}{3}} \right) = - \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) = - \frac{3}{8}\end{array}\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 8 2020
a/
\(sin^2x-sinx=2\left(1-sin^2x\right)\)
\(\Leftrightarrow3sin^2x-sinx-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\frac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{2}+k2\pi\\x=arcsin\left(\frac{2}{3}\right)+k2\pi\\x=\pi-arcsin\left(\frac{2}{3}\right)+k2\pi\end{matrix}\right.\)
2.
\(2sin^2x+\left(1-\sqrt{3}\right)sinx-\frac{\sqrt{3}}{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=\frac{\sqrt{3}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=\frac{\pi}{3}+k2\pi\\x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
3.
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{4}=\frac{\pi}{8}+k2\pi\\3x+\frac{\pi}{4}=-\frac{\pi}{8}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{24}+\frac{k2\pi}{3}\\x=-\frac{\pi}{8}+\frac{k2\pi}{3}\end{matrix}\right.\)
\(\Leftrightarrow\frac{\sqrt{2}}{2}\sin\frac{x}{2}-\frac{\sqrt{2}}{2}\cos\frac{x}{3}=0\)
\(\Leftrightarrow\sin\frac{x}{2}-\cos\frac{x}{3}=0\)
\(\Leftrightarrow\sin\frac{3x}{6}-\cos\frac{2x}{6}=0\)
\(\Leftrightarrow3\sin\frac{x}{6}-4\sin^3\frac{x}{6}-1+2\sin^2\frac{x}{6}=0\) Đặt \(\sin\frac{x}{6}=t\)
\(\Rightarrow4t^3-2t^2-3t+1=0\)
\(\Leftrightarrow\left(t-1\right)\left(4t^2+2t-1\right)=0\)
Giải phương trình tích trên tìm t từ đó suy ra x. Bạn tự làm nốt nhé