Giải phương trình: \(\left(\sin x-2\cos x\right)\cos2x+\sin x=\left(\cos4x-1\right)\cos x+\frac{\cos2x}{2\sin x}\)
Những câu hỏi liên quan
Giải PT
a1) \(3.\cos4x-2^{ }\cos^23x=1\)
a2) \(2\cos2x-8\cos x+7=\dfrac{1}{\cos x}\)
a3) \(\dfrac{\left(1+\sin x+\cos2x\right)\sin\left(x+\dfrac{\pi}{4}\right)}{1+\tan x}=\dfrac{1}{\sqrt{2}}\cos x\)
a4) \(9\sin x+6\cos x-3\sin2x+\cos2x=8\)
a) Pt \(\Leftrightarrow3.cos4x-\left(cos6x+1\right)=1\)
\(\Leftrightarrow3cos4x-cos6x-2=0\)
Đặt \(t=2x\)
Pttt:\(3cos2t-cos3t-2=0\)
\(\Leftrightarrow3\left(2cos^2t-1\right)-\left(4cos^3t-3cost\right)-2=0\)
\(\Leftrightarrow-4cos^3t+6cos^2t+3cost-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cost=1\\cost=\dfrac{1+\sqrt{21}}{4}\left(vn\right)\\cost=\dfrac{1-\sqrt{21}}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=k2\pi\\t=\pm arc.cos\left(\dfrac{1-\sqrt{21}}{4}\right)+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{1}{2}.arccos\left(\dfrac{1-\sqrt{21}}{4}\right)+k\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
a2) \(2cos2x-8cosx+7=\dfrac{1}{cosx}\) (ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\))
\(\Leftrightarrow2.\left(2cos^2x-1\right)-8cosx+7=\dfrac{1}{cosx}\)
\(\Leftrightarrow2.\left(2cos^2x-1\right)cosx-8cos^2x+7cosx=1\)
\(\Leftrightarrow4cos^3x-8cos^2x+5cosx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) (tm) (\(k\in Z\))
Vậy...
a3) Đk: \(x\ne-\dfrac{\pi}{4}+k\pi;x\ne\dfrac{\pi}{2}+k\pi\)
Pt \(\Leftrightarrow\dfrac{\left(1+sinx+1-2sin^2x\right).\dfrac{1}{\sqrt{2}}\left(sinx+cosx\right)}{1+\dfrac{sinx}{cosx}}=\dfrac{1}{\sqrt{2}}cosx\)
\(\Leftrightarrow\dfrac{\left(-2sin^2x+sinx+2\right).\left(sinx+cosx\right)cosx}{cosx+sinx}=cosx\)
\(\Leftrightarrow\left(2+sinx-2sin^2x\right).cosx=cosx\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\2+sinx-2sin^2x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
Đúng 1
Bình luận (2)
a4) Pt \(\Leftrightarrow9sinx+6cosx-6sinx.cosx+1-2sin^2x=8\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sin^2x-9sinx+7\right)=0\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sinx-7\right)\left(sinx-1\right)=0\)
\(\Leftrightarrow\left(1-sinx\right)\left(6cosx+2sinx+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\6cosx+2sinx=7\left(vn\right)\end{matrix}\right.\) (\(6cosx+2sinx=7\) vô nghiệm do \(6^2+2^2< 7^2\))
\(\Rightarrow sinx=1\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi;k\in Z\)
Vậy...
Đúng 1
Bình luận (0)
giải phương trình
sin xsqrt{1+2sin x}cos2x
sinleft(frac{5x}{2}-frac{pi}{4}right)-cosleft(frac{x}{2}-frac{pi}{4}right)sqrt{2}cosfrac{3x}{2}
3sqrt{tan x+1}left(sin x+2cos xright)5left(sin x+3cos xright)
sqrt{2}left(sin x+sqrt{3}cos xright)sqrt{3}cos2x-sin2x
sin2xsin4x+2left(3sin x-4sin^2x+1right)0
Đọc tiếp
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 \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
a/sin3x+cos2x1+2sin xcos2xb/sin^3x+cos^3x2left(sin^5x+cos^5xright)c/dfrac{tan x}{sin x}-dfrac{sin x}{cos x}dfrac{sqrt{2}}{2}d/dfrac{cos xleft(cos x+2sin xright)+3sin xleft(sin x+sqrt{2}right)}{sin2x-1}1e/sin^2x+sin^23x-2cos^22x0f/dfrac{tan x-sin x}{sin^3x}dfrac{1}{cos x}g/sin2xleft(cos x+tan2xright)4cos^2xh/sin^2x+sin^23xcos^2x+cos^23xk/4sin2xdfrac{cos^2x-sin^2x}{cos^6x+sin^6x}mọi người giải giúp em với em đang cần gấp ạ
Đọc tiếp
a/\(\sin3x+\cos2x=1+2\sin x\cos2x\)
b/\(\sin^3x+\cos^3x=2\left(\sin^5x+\cos^5x\right)\)
c/\(\dfrac{\tan x}{\sin x}-\dfrac{\sin x}{\cos x}=\dfrac{\sqrt{2}}{2}\)
d/\(\dfrac{\cos x\left(\cos x+2\sin x\right)+3\sin x\left(\sin x+\sqrt{2}\right)}{\sin2x-1}=1\)
e/\(\sin^2x+\sin^23x-2\cos^22x=0\)
f/\(\dfrac{\tan x-\sin x}{\sin^3x}=\dfrac{1}{\cos x}\)
g/\(\sin2x\left(\cos x+\tan2x\right)=4\cos^2x\)
h/\(\sin^2x+\sin^23x=\cos^2x+\cos^23x\)
k/\(4\sin2x=\dfrac{\cos^2x-\sin^2x}{\cos^6x+\sin^6x}\)
mọi người giải giúp em với em đang cần gấp ạ
Giải phương trình
\(\left(2\cos x+\sqrt{3}\right)\left(\cos2x+2\sin x-\sqrt{3}\right)=1-4\sin^2x\)
\(\left(2cosx+\sqrt{3}\right)\left(cos2x+2sinx-\sqrt{3}\right)=1-4\left(1-cos^2x\right)\)
\(\Leftrightarrow\left(2cosx+\sqrt{3}\right)\left(cos2x+2sinx-\sqrt{3}\right)=4cos^2x-3\)
\(\Leftrightarrow\left(2cosx+\sqrt{3}\right)\left(cos2x+2sinx-\sqrt{3}\right)=\left(2cosx+\sqrt{3}\right)\left(2cosx-\sqrt{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\dfrac{\sqrt{3}}{2}\Rightarrow x=...\\cos2x+2sinx-\sqrt{3}=2cosx-\sqrt{3}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow cos^2x-sin^2x-2\left(cosx-sinx\right)=0\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+sinx\right)-2\left(cosx-sinx\right)=0\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+sinx-2\right)=0\)
\(\Leftrightarrow...\)
Đúng 0
Bình luận (0)
6 tháng 12 2016 lúc 22:44
giải các phương trình sau :a) sinleft(x-frac{2pi}{3}right)cos2x ; b) tanleft(2x+45^oright)tanleft(180^o-frac{x}{2}right)1 ; c) cos2x-sin^2x0 ; d) 5tan x-2cot x3 ; e) sin2x+sin^2xfrac{1}{2} ; f) sin^2frac{x}{2}+sin x-2cos^2frac{x}{2}frac{1}{2} ; g) frac{1+cos2x}{cos x}frac{sin2x}{1-cos2x}
Đọc tiếp
giải các phương trình sau :
a) \(\sin\left(x-\frac{2\pi}{3}\right)=\cos2x\) ; b) \(\tan\left(2x+45^o\right)\tan\left(180^o-\frac{x}{2}\right)=1\) ; c) \(\cos2x-\sin^2x=0\) ; d) \(5\tan x-2\cot x=3\) ; e)
\(\sin2x+\sin^2x=\frac{1}{2}\) ; f) \(\sin^2\frac{x}{2}+\sin x-2\cos^2\frac{x}{2}=\frac{1}{2}\) ; g) \(\frac{1+\cos2x}{\cos x}=\frac{\sin2x}{1-\cos2x}\)
mai đăng lại bài này nhé t làm cho h đi ngủ
Đúng 0
Bình luận (1)
Giải các PT sau:1. dfrac{left(2cos2x-1right)left(sin x-3right)}{sin x}0 2.dfrac{3left(sin x+cos xright)}{sin x-cos x}2+2cos x3.dfrac{3left(sin x+tan xright)}{tan x-sin x}-2cos x24. 1+sin x+cos x+sin2x+cos2x05. 2sin xleft(1+cos2xright)+sin2x1+2cos x
Đọc tiếp
Giải các PT sau:
1. \(\dfrac{\left(2\cos2x-1\right)\left(\sin x-3\right)}{\sin x}=0\)
2.\(\dfrac{3\left(\sin x+\cos x\right)}{\sin x-\cos x}=2+2\cos x\)
3.\(\dfrac{3\left(\sin x+\tan x\right)}{\tan x-\sin x}-2\cos x=2\)
4. \(1+\sin x+\cos x+\sin2x+\cos2x=0\)
5. \(2\sin x\left(1+\cos2x\right)+\sin2x=1+2\cos x\)
1.
ĐKXĐ: \(x\ne k\pi\)
\(\Leftrightarrow\left(2cos2x-1\right)\left(sinx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{1}{2}\\sinx=3>1\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{3}+k2\pi\\2x=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
Đúng 1
Bình luận (0)
2. Bạn kiểm tra lại đề, pt này về cơ bản ko giải được.
3.
ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)
\(\dfrac{3\left(sinx+\dfrac{sinx}{cosx}\right)}{\dfrac{sinx}{cosx}-sinx}-2cosx=2\)
\(\Leftrightarrow\dfrac{3\left(1+cosx\right)}{1-cosx}+2\left(1+cosx\right)=0\)
\(\Leftrightarrow\left(1+cosx\right)\left(\dfrac{3}{1-cosx}+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\left(loại\right)\\cosx=\dfrac{5}{2}\left(loại\right)\end{matrix}\right.\)
Vậy pt đã cho vô nghiệm
Đúng 1
Bình luận (1)
4.
\(\Leftrightarrow\left(sin^2x+cos^2x+2sinx.cosx\right)+\left(sinx+cosx\right)+\left(cos^2x-sin^2x\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)^2+\left(sinx+cosx\right)+\left(sinx+cosx\right)\left(cosx-sinx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sinx+cosx+1+cosx-sinx\right)=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{2\pi}{3}+k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
Đúng 1
Bình luận (0)
Xem thêm câu trả lời
chứng minh các đẳng thức sau
a) \(\cos x\cos\left(\frac{\pi}{3}-x\right)\cos\left(\frac{\pi}{3}+x\right)=\frac{1}{4}\cos3x\)
b) \(\sin5x-2\sin x\left(\cos4x+\cos2x\right)=\sin x\)
\(cosx.cos\left(\frac{\pi}{3}-x\right)cos\left(\frac{\pi}{3}+x\right)=\frac{1}{2}cosx\left(cos\frac{2\pi}{3}+cos2x\right)=-\frac{1}{4}cosx+\frac{1}{2}cosx.cos2x\)
\(=-\frac{1}{4}cosx+\frac{1}{4}\left(cos3x+cosx\right)=\frac{1}{4}cos3x\)
\(sin5x-2sinx\left(cos4x+cos2x\right)=sinx.cos4x+cosx.sin4x-2sinx.cos4x-2sinx.cos2x\)
\(=sin4x.cosx-cos4x.sinx-2sinx.cos2x=sin3x-2sinx.cos2x\)
\(=sinx.cos2x+cosx.sin2x-2sinx.cos2x\)
\(=sin2x.cosx-cos2x.sinx=sinx\)
Đúng 0
Bình luận (0)
Giải phương trình: \(\left(2\cos2x-1\right)\cos x-\sin x=\sqrt{2}\left(\sin x+\cos x\right)\sin3x\)
\(\Leftrightarrow2cosx.cos2x-\left(cosx+sinx\right)-\sqrt{2}sin3x\left(cosx+sinx\right)=0\)
\(\Leftrightarrow2cosx\left(cos^2x-sin^2x\right)-\left(cosx+sinx\right)\left(1+\sqrt{2}sin3x\right)=0\)
\(\Leftrightarrow\left(cosx+sinx\right)\left(2cos^2x-2sinx.cosx\right)-\left(cosx+sinx\right)\left(1+\sqrt{2}sin3x\right)=0\)
\(\Leftrightarrow\left(cosx+sinx\right)\left(2cos^2x-sin2x-1-\sqrt{2}sin3x\right)=0\)
Biến đổi ngoặc sau:
\(cos2x-sin2x=\sqrt{2}sin3x\)
\(\Leftrightarrow-\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}sin3x\)
\(\Leftrightarrow sin\left(\frac{\pi}{4}-2x\right)=sin3x\)
Đúng 0
Bình luận (0)
Giải các phương trình sau:
\(5\sin^22x-6\sin4x-2\cos^2x=0\)
\(2\sin^23x-10\sin6x-\cos^23x=-2\)
\(\sin^2x\left(\tan x+1\right)=3\sin x\left(\cos x-\sin x\right)+3\)
\(6\sin x-2\cos^3x=\frac{5\sin4x.\cos x}{2\cos2x}\)