\(\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\cos x}}}\)=\(\cos\frac{x}{n}\) 0<x<\(\frac{\Pi}{2}\)
với giá trị nào của n thì đẳng thức sau luôn đúng
\(\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cosx}}}\) ta được M = cos\(\frac{x}{n}\) (0<x< \(\frac{\pi}{2}\)).Tìm n
\(M=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{x}{2}-1\right)}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{cos^2\frac{x}{2}}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{x}{2}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{cos^2\frac{x}{4}}}\) (tách tương tự như trên)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{x}{4}}=\sqrt{cos^2\frac{x}{8}}=cos\frac{x}{8}\)
\(\Rightarrow n=8\)
Giải các phương trình sau:
a) \(\sin \left( {2x - \frac{\pi }{6}} \right) = - \frac{{\sqrt 3 }}{2}\)
b) \(\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\)
c) \(\sin 3x - \cos 5x = 0\)
d) \({\cos ^2}x = \frac{1}{4}\)
e) \(\sin x - \sqrt 3 \cos x = 0\)
f) \(\sin x + \cos x = 0\)
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)
\(\begin{array}{l}\sin 3x - \cos 5x = 0\\ \Leftrightarrow \sin 3x = \cos 5x\\ \Leftrightarrow \cos 5x = \cos \left( {\frac{\pi }{2} - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} - 3x + k2\pi \\5x = - \left( {\frac{\pi }{2} - 3x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}8x = \frac{\pi }{2} + k2\pi \\2x = - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{16}} + \frac{{k\pi }}{4}\\x = - \frac{\pi }{4} + k\pi \end{array} \right.\end{array}\)
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)
\(\begin{array}{l}\sin x + \cos x = 0\\ \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin x + \frac{{\sqrt 2 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{4}.\sin x + \sin \frac{\pi }{4}.\cos x = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin 0\\ \Leftrightarrow x + \frac{\pi }{4} = k\pi ;k \in Z\\ \Leftrightarrow x = - \frac{\pi }{4} + k\pi ;k \in Z\end{array}\)
1) \(sin^2\left(\frac{x}{2}-\frac{\pi}{4}\right).tan^2x-cos^2\frac{x}{2}=0\)
2) \(tanx=sin^2x\left(c-\frac{\pi}{2010}\right)+cos^2\left(2x+\frac{\pi}{2010}\right)+sinx.sin\left(3x+\frac{\pi}{1005}\right)\)
3) \(1+2cosx\left(sinx-1\right)+\sqrt{2}sinx+4cosx.sin^2\frac{x}{2}=0\)
4) \(3cos4x-8cos^6x+2cos4x=3\)
5) \(1+sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)\)
6) \(sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-4\sqrt{3}cos^2x.sinx.cos2x\)
7) \(\frac{tan^2x+tanx}{tan^2x+1}=\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{4}\right)\)
8) \(cos^4x+sin^4x+cos\left(x-\frac{\pi}{4}\right).sin\left(3x-\frac{\pi}{4}\right)-\frac{3}{2}=0\)
Câu 2 bạn coi lại đề
3.
\(1+2sinx.cosx-2cosx+\sqrt{2}sinx+2cosx\left(1-cosx\right)=0\)
\(\Leftrightarrow sin2x-\left(2cos^2x-1\right)+\sqrt{2}sinx=0\)
\(\Leftrightarrow sin2x-cos2x=-\sqrt{2}sinx\)
\(\Leftrightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}sin\left(-x\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=sin\left(-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-x+k2\pi\\2x-\frac{\pi}{4}=\pi+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
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):
\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)
\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)
\(\Leftrightarrow2sin^3x-sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)
\(\Leftrightarrow...\)
6.
\(sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-2\sqrt{3}cosx.sin2x.cos2x\)
\(\Leftrightarrow sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-\sqrt{3}cosx.sin4x\)
\(\Leftrightarrow sin4x\left(sinx+\sqrt{3}cosx\right)=\sqrt{2}sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow sin4x\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{3}\right)=0\)
\(\Leftrightarrow sin4x.sin\left(x+\frac{\pi}{3}\right)-\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{3}\right)=0\)
\(\Leftrightarrow\left(sin4x-\frac{\sqrt{2}}{2}\right)sin\left(x+\frac{\pi}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin4x=\frac{\sqrt{2}}{2}\\sin\left(x+\frac{\pi}{3}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
Giải các phương trình sau:
a) \(\sin x = \frac{{\sqrt 3 }}{2}\);
b) \(2\cos x = - \sqrt 2 \);
c) \(\sqrt 3 \tan \left( {\frac{x}{2} + {{15}^0}} \right) = 1\);
d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\)
a) \(\sin x = \frac{{\sqrt 3 }}{2}\;\; \Leftrightarrow \sin x = \sin \frac{\pi }{3}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \pi - \frac{\pi }{3} + k2\pi }\end{array}} \right.\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \frac{{2\pi }}{3} + k2\pi \;}\end{array}\;} \right.\left( {k \in \mathbb{Z}} \right)\)
b) \(2\cos x = - \sqrt 2 \;\; \Leftrightarrow \cos x = - \frac{{\sqrt 2 }}{2}\;\;\; \Leftrightarrow \cos x = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{{3\pi }}{4} + k2\pi }\\{x = - \frac{{3\pi }}{4} + k2\pi }\end{array}\;\;\left( {k \in \mathbb{Z}} \right)} \right.\)
c) \(\sqrt 3 \;\left( {\tan \frac{x}{2} + {{15}^0}} \right) = 1\;\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \frac{1}{{\sqrt 3 }}\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \tan \frac{\pi }{6}\)
\( \Leftrightarrow \frac{x}{2} + \frac{\pi }{{12}} = \frac{\pi }{6} + k\pi \;\;\;\; \Leftrightarrow \frac{x}{2} = \frac{\pi }{{12}} + k\pi \;\;\; \Leftrightarrow x = \frac{\pi }{6} + k\pi \;\left( {k \in \mathbb{Z}} \right)\)
d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\;\;\;\; \Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \;\;\;\; \Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \;\; \Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2}\;\;\left( {k \in \mathbb{Z}} \right)\)
giải phương trình
\(\sin x\sqrt{1+2\sin x}=\cos2x\)
\(\sin\left(\frac{5x}{2}-\frac{\pi}{4}\right)-\cos\left(\frac{x}{2}-\frac{\pi}{4}\right)=\sqrt{2}\cos\frac{3x}{2}\)
\(3\sqrt{\tan x+1}\left(\sin x+2\cos x\right)=5\left(\sin x+3\cos x\right)\)
\(\sqrt{2}\left(\sin x+\sqrt{3}\cos x\right)=\sqrt{3}\cos2x-\sin2x\)
\(\sin2x\sin4x+2\left(3\sin x-4\sin^2x+1\right)=0\)
a/ Hmm, bạn có nhầm lẫn chỗ nào ko nhỉ, nghiệm của pt này xấu khủng khiếp
b/ \(\Leftrightarrow sin\frac{5x}{2}-cos\frac{5x}{2}-sin\frac{x}{2}-cos\frac{x}{2}=cos\frac{3x}{2}\)
\(\Leftrightarrow2cos\frac{3x}{2}.sinx-2cos\frac{3x}{2}cosx=cos\frac{3x}{2}\)
\(\Leftrightarrow cos\frac{3x}{2}\left(2sinx-2cosx-1\right)=0\)
\(\Leftrightarrow cos\frac{3x}{2}\left(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)-1\right)=0\)
c/ Do \(cosx\ne0\), chia 2 vế cho cosx ta được:
\(3\sqrt{tanx+1}\left(tanx+2\right)=5\left(tanx+3\right)\)
Đặt \(\sqrt{tanx+1}=t\ge0\)
\(\Leftrightarrow3t\left(t^2+1\right)=5\left(t^2+2\right)\)
\(\Leftrightarrow3t^3-5t^2+3t-10=0\)
\(\Leftrightarrow\left(t-2\right)\left(3t^2+t+5\right)=0\)
d/ \(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{3}\right)=-sin\left(2x-\frac{\pi}{3}\right)\)
Đặt \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
Chứng minh rằng: (Pls help me)
a, \(\frac{1}{\sin x}+\cot x=\cot\frac{x}{2}\)
b, \(\frac{1-\cos x}{\sin x}=\tan\frac{x}{2}\)
c,\(\tan\frac{x}{2}\left(\frac{1}{\cos x}+1\right)=\tan x\)
d,\(\frac{\sin2a}{2\cos a\left(1+\cos a\right)}=\tan\frac{a}{2}\)
e,\(\cot x+\tan\frac{x}{2}=\frac{1}{\sin x}\)
f,\(3-4\cos2x+\cos4x=8\sin^4x\)
g,\(\frac{1-\cos x}{\sin x}=\frac{\sin x}{1+\cos x}\)
h,\(\sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\)
i,\(\sin x-\cos x=\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)\)
l,\(\cos x-\sin x=\sqrt{2}\cos\left(x+\frac{\pi}{4}\right)\)
a/
\(\frac{1}{sinx}+\frac{cosx}{sinx}=\frac{1+cosx}{sinx}=\frac{1+2cos^2\frac{x}{2}-1}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2cos^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{cos\frac{x}{2}}{sin\frac{x}{2}}=cot\frac{x}{2}\)
b/
\(\frac{1-cosx}{sinx}=\frac{1-\left(1-2sin^2\frac{x}{2}\right)}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2sin^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=tan\frac{x}{2}\)
c/
\(tan\frac{x}{2}\left(\frac{1}{cosx}+1\right)=\left(\frac{1-cosx}{sinx}\right)\left(\frac{1}{cosx}+1\right)=\frac{\left(1-cosx\right)\left(1+cosx\right)}{sinx.cosx}=\frac{1-cos^2x}{sinx.cosx}\)
\(=\frac{sin^2x}{sinx.cosx}=\frac{sinx}{cosx}=tanx\)
d/
\(\frac{sin2a}{2cosa\left(1+cosa\right)}=\frac{2sina.cosa}{2cosa\left(1+2cos^2\frac{a}{2}-1\right)}=\frac{sina}{2cos^2\frac{a}{2}}=\frac{2sin\frac{a}{2}cos\frac{a}{2}}{2cos^2\frac{a}{2}}=tan\frac{a}{2}\)
e/
\(cotx+tan\frac{x}{2}=\frac{cosx}{sin}+\frac{1-cosx}{sinx}=\frac{cosx+1-cosx}{sinx}=\frac{1}{sinx}\)
Các câu c, e đều sử dụng kết quả từ câu b
f/
\(3-4cos2x+cos4x=3-4cos2x+2cos^22x-1\)
\(=2cos^22x-4cos2x+2=2\left(cos^22x-2cos2x+1\right)\)
\(=2\left(cos2x-1\right)^2=2\left(1-2sin^2x-1\right)^2\)
\(=2.\left(-2sin^2x\right)^2=8sin^4x\)
g/
\(\frac{1-cosx}{sinx}=\frac{sinx\left(1-cosx\right)}{sin^2x}=\frac{sinx\left(1-cosx\right)}{1-cos^2x}=\frac{sinx\left(1-cosx\right)}{\left(1-cosx\right)\left(1+cosx\right)}=\frac{sinx}{1+cosx}\)
h/
\(sinx+cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}+cosx.\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
i/
\(sinx-cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}-cosx.\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\)
j/
\(cosx-sinx=\sqrt{2}\left(cosx.\frac{\sqrt{2}}{2}-sinx\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(cosx.cos\frac{\pi}{4}-sinx.sin\frac{\pi}{4}\right)=\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)
Đưa về tích rồi giải các phương trình sau:
a) \(\sin 2x -2.\sin x +\cos x -1=0\)
b) \(\sqrt{2} . (\sin x - 2.\cos x) = 2-\sin 2x\)
c) \(\frac{1}{\cos x} - \frac{1}{\sin x}=2\sqrt 2 .\cos(x + \frac{\pi}{4}) \)
\(a,sin2x-2sinx+cosx-1=0\)
\(\Leftrightarrow2sinxcosx-2sinx+cosx-1=0\)
\(\Leftrightarrow2sinx\left(cosx-1\right)+cosx-1=0\)
\(\Leftrightarrow\left(cosx-1\right)\left(2sinx+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}cosx=1\\sinx=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2k\pi\\x=\frac{-\pi}{6}+2k\pi\end{cases}}}\)
\(b,\sqrt{2}\left(sinx-2cosx\right)=2-sin2x\)
\(\Leftrightarrow\sqrt{2}sinx-2\sqrt{2}cosx-2+2sinxcosx=0\)
\(\Leftrightarrow\sqrt{2}sinx\left(1+\sqrt{2}cosx\right)-2.\left(\sqrt{2}cosx+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{2}cosx+1\right)\left(\sqrt{2}sinx-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}cosx=\frac{-\sqrt{2}}{2}\\sinx=\frac{2\sqrt{2}}{2}\left(l\right)\end{cases}}\)(vì \(-1\le sinx\le1\))
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3\pi}{4}+2k\pi\\x=\frac{5\pi}{4}+2k\pi\end{cases}}\)
\(c,\frac{1}{cosx}-\frac{1}{sinx}=2\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)
\(\Leftrightarrow\frac{sinx-cosx}{sinx.cosx}=2\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)
\(\Leftrightarrow\frac{-\sqrt{2}cos\left(x+\frac{\pi}{4}\right)}{sinx.cosx}=2\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)
\(\Leftrightarrow sin2x+1=0\)
\(\Leftrightarrow sin2x=-1\)
\(\Leftrightarrow2x=\frac{3\pi}{2}+2k\pi\)
\(\Leftrightarrow x=\frac{3\pi}{4}+k\pi\)
@Bùi Nhật Vy, Bạn nhớ kĩ cái này nha
\(asinx+bcosx=\sqrt{a^2+b^2}sin\left(x+\alpha\right)=-\sqrt{a^2+b^2}cos\left(x-\alpha\right)\)
trong đó \(\cos\alpha=\frac{a}{\sqrt{a^2+b^2}},sin\alpha=\frac{b}{\sqrt{a^2+b^2}}\)
Giải các phương trình sau
1) \(\cos^5\frac{x}{2}\)\(\sin\frac{x}{2}\) - \(\sin^5\frac{x}{2}\)\(\cos\frac{x}{2}\) = \(\frac{\sqrt{3}}{8}\)
2) 8\(\cos x.\cos\left(\frac{\pi}{3}+x\right).\cos\left(\frac{\pi}{3}-x\right)\) = \(\sqrt{3}\)
3) \(\frac{1}{\cos x}+\frac{1}{\sin2x}=\frac{2}{\sin4x}\)
1.
\(\Leftrightarrow2sin\frac{x}{2}cos\frac{x}{2}\left(cos^4\frac{x}{2}-sin^4\frac{x}{2}\right)=\frac{\sqrt{3}}{4}\)
\(\Leftrightarrow sinx\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\left(cos^2\frac{x}{2}+sin^2\frac{x}{2}\right)=\frac{\sqrt{3}}{4}\)
\(\Leftrightarrow sinx.cosx=\frac{\sqrt{3}}{4}\)
\(\Leftrightarrow\frac{1}{2}sin2x=\frac{\sqrt{3}}{4}\)
\(\Leftrightarrow sin2x=\frac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{3}+k2\pi\\2x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=\frac{\pi}{3}+k\pi\end{matrix}\right.\)
2.
\(\Leftrightarrow4cosx\left(cos\frac{2\pi}{3}+cos2x\right)=\sqrt{3}\)
\(\Leftrightarrow4cosx\left(cos2x-\frac{1}{2}\right)=\sqrt{3}\)
\(\Leftrightarrow4cos2x.cosx-2cosx=\sqrt{3}\)
\(\Leftrightarrow2cos3x+2cosx-2cosx=\sqrt{3}\)
\(\Leftrightarrow cos3x=\frac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\frac{\pi}{6}+k2\pi\\3x=-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{18}+\frac{k2\pi}{3}\\x=-\frac{\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
3.
ĐKXĐ: ...
\(\frac{1}{cosx}+\frac{1}{2sinx.cosx}=\frac{1}{2sinx.cosx.cos2x}\)
\(\Leftrightarrow2sinx.cos2x+cos2x=1\)
\(\Leftrightarrow2sinx.cos2x+1-2sin^2x=1\)
\(\Leftrightarrow2sinx\left(cos2x-sinx\right)=0\)
\(\Leftrightarrow cos2x-sinx=0\)
\(\Leftrightarrow1-2sin^2x-sinx=0\)
\(\Leftrightarrow2sin^2x+sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
A=\(\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\cdot cosa}}}\) với 0<a<bi/2.rút gọn biểu thức A ta được A=cos\(\frac{a}{n}\)hãy cho biết n thuộc khoảng nào
\(A=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cosa}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{2}-1\right)}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+cos^2\frac{a}{2}-\frac{1}{2}}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{a}{2}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{4}-1\right)}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{a}{4}}=\sqrt{\frac{1}{2}+\frac{1}{2}\left(cos^2\frac{a}{8}-1\right)}\)
\(=cos\frac{a}{8}\Rightarrow n=8\)