Giải phương trình sau :
a) \(1+\sin x-\cos x-\sin2x+2\cos2x=0\)
b) \(\sin x-\dfrac{1}{\sin x}=\sin^2x-\dfrac{1}{\sin^2x}\)
c) \(\cos x\tan3x=\sin5x\)
d) \(2\tan^2x+3\tan x+2cot^2x+3cotx+2=0\)
giải pt
a, \(\sin^2x+\sin^22x+\sin^23x=\dfrac{3}{2}\)
b. \(\cos^2x+\sin^22x+\cos^23x=1\)
c,\(\sin5x+2\cos^2x=1\)
d,\(1+\tan x=2\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)\)
e,\(\sin3x+\cos3x-\sin x+\cos x=\sqrt{2}\cos2x\)
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 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: \(x\ne k\pi\)
\(\Leftrightarrow\left(2cos2x-1\right)\left(sinx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{1}{2}\\sinx=3>1\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{3}+k2\pi\\2x=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
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
4.
\(\Leftrightarrow\left(sin^2x+cos^2x+2sinx.cosx\right)+\left(sinx+cosx\right)+\left(cos^2x-sin^2x\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)^2+\left(sinx+cosx\right)+\left(sinx+cosx\right)\left(cosx-sinx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sinx+cosx+1+cosx-sinx\right)=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{2\pi}{3}+k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
Giải các phương trình
a) \(\dfrac{\cos2x}{\sin2x-1}=0\)
b) \(\cos\left(\sin x\right)=1\)
c) \(2\sin^2x-1+\cos3x=0\)
d) \(tan3x.tanx=1\)
e) \(\cos3x=-\cos7x\)
a: ĐKXĐ: sin 2x<>1
=>2x<>pi/2+k2pi
=>x<>pi/4+kpi
\(\dfrac{cos2x}{sin2x-1}=0\)
=>cos2x=0
=>2x=pi/2+kpi
=>x=pi/4+kpi/2
Kết hợp ĐKXĐ, ta được:
x=3/4pi+k2pi hoặc x=7/4pi+k2pi
b: cos(sinx)=1
=>sin x=kpi
=>sin x=0
=>x=kpi
c: \(2\cdot sin^2x-1+cos3x=0\)
=>cos3x+cos2x=0
=>cos3x=-cos2x=-sin(pi/2-2x)=sin(2x-pi/2)
=>cos3x=cos(pi/2-2x+pi/2)=cos(pi-2x)
=>3x=pi-2x+k2pi hoặc 3x=-pi+2x+k2pi
=>x=-pi+k2pi hoặc x=pi/5+k2pi/5
e: cos3x=-cos7x
=>cos3x=cos(pi-7x)
=>3x=pi-7x+k2pi hoặc 3x=-pi+7x+k2pi
=>x=pi/10+kpi/5 hoặc x=pi/4-kpi/2
Chứng minh:
1.\(\dfrac{\cot^2x-\sin^2x}{\cot^2x-\tan^2x}=\sin^2x\cdot\cos^2x\)
2.\(\dfrac{1-\sin x}{\cos x}-\dfrac{\cos x}{1+\sin x}=0\)
3.\(\dfrac{\tan x}{\sin x}-\dfrac{\sin x}{\cot x}=\cos x\)
4.\(\dfrac{\tan x}{1-\tan^2x}\cdot\dfrac{\cot^2x-1}{\cot x}=1\)
5.\(\dfrac{1+\sin^2x}{1-\sin^2x}=1+2\tan^2x\)
Câu 1 đề sai, chắc chắn 1 trong 2 cái \(cot^2x\) phải có 1 cái là \(cos^2x\)
2.
\(\dfrac{1-sinx}{cosx}-\dfrac{cosx}{1+sinx}=\dfrac{\left(1-sinx\right)\left(1+sinx\right)-cos^2x}{cosx\left(1+sinx\right)}=\dfrac{1-sin^2x-cos^2x}{cosx\left(1+sinx\right)}\)
\(=\dfrac{1-\left(sin^2x+cos^2x\right)}{cosx\left(1+sinx\right)}=\dfrac{1-1}{cosx\left(1+sinx\right)}=0\)
3.
\(\dfrac{tanx}{sinx}-\dfrac{sinx}{cotx}=\dfrac{tanx.cotx-sin^2x}{sinx.cotx}=\dfrac{1-sin^2x}{sinx.\dfrac{cosx}{sinx}}=\dfrac{cos^2x}{cosx}=cosx\)
4.
\(\dfrac{tanx}{1-tan^2x}.\dfrac{cot^2x-1}{cotx}=\dfrac{tanx}{1-tan^2x}.\dfrac{\dfrac{1}{tan^2x}-1}{\dfrac{1}{tanx}}=\dfrac{tanx}{1-tan^2x}.\dfrac{1-tan^2x}{tanx}=1\)
5.
\(\dfrac{1+sin^2x}{1-sin^2x}=\dfrac{1+sin^2x}{cos^2x}=\dfrac{1}{cos^2x}+tan^2x=\dfrac{sin^2x+cos^2x}{cos^2x}+tan^2x\)
\(=tan^2x+1+tan^2x=1+2tan^2x\)
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
Giải các phương trình :
a) \(\cos^2x+\cos^22x-\cos^23x-\cos^24x=0\)
b) \(\cos4x\cos\left(\pi+2x\right)-\sin2x\cos\left(\dfrac{\pi}{2}-4x\right)=\dfrac{\sqrt{2}}{2}\sin4x\)
c) \(\tan\left(120^0+3x\right)-\tan\left(140^0-x\right)=2\sin\left(80^0+2x\right)\)
d) \(\tan^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}\tan\dfrac{x}{2}+\cos^2\dfrac{x}{2}+\cot^2\dfrac{x}{2}+\sin x=4\)
e) \(\dfrac{\sin2t+2\cos^2t-1}{\cot t-\cot3t+\sin3t-\sin t}=\cos t\)
Giải các phương trình sau
a) \(sin^6x+cos^6x=cos2x+\dfrac{1}{16}\)
b) \(sin^4\dfrac{x}{2}+cos^4\dfrac{x}{2}=\dfrac{5}{2}-2sinx\)
c) \(cos5xcosx=cos4xcos2x+4-3sin^2x\)
d) \(2cosxcos2x=1+cos2x+cos3x\)
e) \(sin3x+cos2x=2\left(sin2xcosx-1\right)\)
a.
\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos2x+\dfrac{1}{16}\)
\(\Leftrightarrow1-\dfrac{3}{4}sin^22x=cos2x+\dfrac{1}{16}\)
\(\Leftrightarrow\dfrac{15}{16}-\dfrac{3}{4}\left(1-cos^22x\right)=cos2x\)
\(\Leftrightarrow\dfrac{3}{4}cos^22x-cos2x+\dfrac{3}{16}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{4-\sqrt{7}}{6}\\cos2x=\dfrac{4+\sqrt{7}}{6}>1\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm\dfrac{1}{2}arccos\left(\dfrac{4-\sqrt{7}}{6}\right)+k\pi\)
b.
\(\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}=\dfrac{5}{2}-2sinx\)
\(\Leftrightarrow1-\dfrac{1}{2}sin^2x=\dfrac{5}{2}-2sinx\)
\(\Leftrightarrow\dfrac{1}{2}sin^2x-2sinx+\dfrac{3}{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=3\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)
c.
\(\Leftrightarrow\dfrac{1}{2}cos6x+\dfrac{1}{2}cos4x=\dfrac{1}{2}cos6x+\dfrac{1}{2}cos2x+4-3\left(\dfrac{1}{2}-\dfrac{1}{2}cos2x\right)\)
\(\Leftrightarrow\dfrac{1}{2}\left(2cos^22x-1\right)=\dfrac{1}{2}cos2x+\dfrac{5}{2}+\dfrac{3}{2}cos2x\)
\(\Leftrightarrow cos^22x-2cos2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=3\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow2x=\pi+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)
1. Tìm m để PT có nghiệm:
a) \(\sqrt{3}\cos^2x+\dfrac{1}{2}\sin2x=m\)
b) \(3\sin^2x-2\sin x\cos x+m=0\)
c) \(\sin^2x+2\left(m-1\right)\sin x\cos x-\left(m+1\right)\cos^2x=m\)
b.
\(\Leftrightarrow\dfrac{3}{2}\left(1-cos2x\right)-sin2x+m=0\)
\(\Leftrightarrow sin2x+\dfrac{3}{2}cos2x-\dfrac{3}{2}=m\)
\(\Leftrightarrow\dfrac{\sqrt{13}}{2}\left(\dfrac{2}{\sqrt{13}}sin2x+\dfrac{3}{\sqrt{13}}cos2x\right)-\dfrac{3}{2}=m\)
Đặt \(\dfrac{2}{\sqrt{13}}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\)
\(\Rightarrow\dfrac{\sqrt{13}}{2}sin\left(2x+a\right)-\dfrac{3}{2}=m\)
Phương trình có nghiệm khi và chỉ khi:
\(\dfrac{-\sqrt{13}-3}{2}\le m\le\dfrac{\sqrt{13}-3}{2}\)
Lý thuyết đồ thị:
Phương trình \(f\left(x\right)=m\) có nghiệm khi và chỉ khi \(f\left(x\right)_{min}\le m\le f\left(x\right)_{max}\)
Hoặc sử dụng điều kiện có nghiệm của pt lương giác bậc nhất (tùy bạn)
a.
\(\dfrac{\sqrt{3}}{2}\left(1-cos2x\right)+\dfrac{1}{2}sin2x=m\)
\(\Leftrightarrow\dfrac{1}{2}sin2x-\dfrac{\sqrt{3}}{2}cos2x+\dfrac{\sqrt{3}}{2}=m\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)+\dfrac{\sqrt{3}}{2}=m\)
\(\Rightarrow\) Pt có nghiệm khi và chỉ khi:
\(-1+\dfrac{\sqrt{3}}{2}\le m\le1+\dfrac{\sqrt{3}}{2}\)
c.
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos2x+\left(m-1\right)sin2x-\left(m+1\right)\left(\dfrac{1}{2}+\dfrac{1}{2}cos2x\right)=m\)
\(\Leftrightarrow\left(2m-2\right)sin2x-\left(m+2\right)cos2x=3m\)
Theo điều kiện có nghiệm của pt lượng giác bậc nhất, pt có nghiệm khi:
\(\left(2m-2\right)^2+\left(m+2\right)^2\ge9m^2\)
\(\Leftrightarrow m^2+m-2\le0\)
\(\Leftrightarrow-2\le m\le\)