Giai phương trình bậc hai theo sin hoặc cos có biến đổi đơn giản như :
a/ \(sin^2x+3cosx-3=0\)
b/ \(2cos^2x+sinx-1=0\)
c/ \(cos2x-5cosx+2=0\)
d/ \(cos2x+2sinx-2=0\)
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Giair phương trình lượng giác sau:
1) cosx - cos2x +cos3x = 0
2) cos2x - sin2x = sin3x + cos4x
3) cos2x + 2sinx - 1 - 2sinxsosx = 0
4) 1+ sinx - cosx = sin2x - cos2x
5) \(\sqrt{2}\) sin (2x+\(\dfrac{\pi}{4}\)) - sinx - 3cosx +2 =0
6) sin2x + 2cos2x = 1+sinx - 4cosx
giải các pt sau:
a) cosx(1-cos2x) - sin^2x = 0
b) sin3x + cos2x = 1 + 2sinxcos3x
c) ( cosx+1)(sinx - cosx + 3) = sin^2x
d) (1+sinx)(cosx-sinx) = cos^2x
a.
\(\Leftrightarrow cosx\left[1-\left(1-2sin^2x\right)\right]-sin^2x=0\)
\(\Leftrightarrow2sin^2x.cosx-sin^2x=0\)
\(\Leftrightarrow sin^2x\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
b.
Câu b chắc chắn đề đúng chứ bạn? Vế phải ấy?
c/
\(\left(1+cosx\right)\left(sinx-cosx+3\right)=1-cos^2x\)
\(\Leftrightarrow\left(1+cosx\right)\left(sinx-cosx+3\right)-\left(1+cosx\right)\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1+cosx\right)\left(sinx+2\right)=0\)
\(\Leftrightarrow cosx=-1\)
\(\Leftrightarrow x=\pi+k2\pi\)
d.
\(\Leftrightarrow\left(1+sinx\right)\left(cosx-sinx\right)=1-sin^2x\)
\(\Leftrightarrow\left(1+sinx\right)\left(cosx-sinx\right)-\left(1+sinx\right)\left(1-sinx\right)=0\)
\(\Leftrightarrow\left(1+sinx\right)\left(cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{2}+k2\pi\\x=k2\pi\end{matrix}\right.\)
giai pt:
a) \(4sin^5x.cosx-4cos^5x.sinx=sin^24x\)
b) \(4sin^2\frac{x}{2}-\sqrt{3}cos2x=1+2cos^2\left(x-\frac{3\pi}{4}\right)\)
c) \(sin^2\left(x+\frac{\pi}{3}\right)+sinx+\sqrt{3}cosx=\frac{5}{4}\)
d) \(2sinx\left(1+cos2x\right)+sin2x=1+2cosx\)
e) \(sin^2x+4sinx.cosx+3cos^2x-sinx-3ccosx=0\)
a/
\(\Leftrightarrow4sinx.cosx\left(sin^4x-cos^4x\right)=sin^24x\)
\(\Leftrightarrow2sin2x\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin^24x\)
\(\Leftrightarrow-2sin2x.cos2x=sin^24x\)
\(\Leftrightarrow-sin4x=sin^24x\)
\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\sin4x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=k\pi\\4x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{4}\\x=-\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)
b/
\(\Leftrightarrow2\left(1-cosx\right)-\sqrt{3}cos2x=1+1+cos\left(2x-\frac{3\pi}{2}\right)\)
\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=sin\left(2\pi-2x\right)\)
\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=-sin2x\)
\(\Leftrightarrow sin2x-\sqrt{3}cos2x=2cosx\)
\(\Leftrightarrow\frac{1}{2}sin2x-\sqrt{3}cos2x=cosx\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=cosx=sin\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{2}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{5}{4}=0\)
\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{3}\right)-\frac{5}{4}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\\sin\left(x+\frac{\pi}{3}\right)=-\frac{5}{2}< -1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
1,Giải phương trình:
a,\(cos^3x+sin^3x=cos2x\)
b,\(cos^3x+sin^3x=2sin2x+sinx+cosx\)
c,\(2cos^3x=sin3x\)
d,\(cos^2x-\sqrt{3}sin2x=1+sin^2x\)
e,\(cos^3x+sin^3x=2\left(cos^5x+sin^5x\right)\)
a, (sinx + cosx)(1 - sinx . cosx) = (cosx - sinx)(cosx + sinx)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx-sinx=1-sinx.cosx\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx+sinx.cosx-1-sinx=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\\left(cosx-1\right)\left(sinx+1\right)=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=1\\sinx=-1\end{matrix}\right.\)
b, (sinx + cosx)(1 - sinx . cosx) = 2sin2x + sinx + cosx
⇔ (sinx + cosx)(1 - sinx.cosx - 1) = 2sin2x
⇔ (sinx + cosx).(- sinx . cosx) = 2sin2x
⇔ 4sin2x + (sinx + cosx) . sin2x = 0
⇔ \(\left[{}\begin{matrix}sin2x=0\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+4=0\end{matrix}\right.\)
⇔ sin2x = 0
c, 2cos3x = sin3x
⇔ 2cos3x = 3sinx - 4sin3x
⇔ 4sin3x + 2cos3x - 3sinx(sin2x + cos2x) = 0
⇔ sin3x + 2cos3x - 3sinx.cos2x = 0
Xét cosx = 0 : thay vào phương trình ta được sinx = 0. Không có cung x nào có cả cos và sin = 0 nên cosx = 0 không thỏa mãn phương trình
Xét cosx ≠ 0 chia cả 2 vế cho cos3x ta được :
tan3x + 2 - 3tanx = 0
⇔ \(\left[{}\begin{matrix}tanx=1\\tanx=-2\end{matrix}\right.\)
d, cos2x - \(\sqrt{3}sin2x\) = 1 + sin2x
⇔ cos2x - sin2x - \(\sqrt{3}sin2x\) = 1
⇔ cos2x - \(\sqrt{3}sin2x\) = 1
⇔ \(2cos\left(2x+\dfrac{\pi}{3}\right)=1\)
⇔ \(cos\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}=cos\dfrac{\pi}{3}\)
e, cos3x + sin3x = 2cos5x + 2sin5x
⇔ cos3x (1 - 2cos2x) + sin3x (1 - 2sin2x) = 0
⇔ cos3x . (- cos2x) + sin3x . cos2x = 0
⇔ \(\left[{}\begin{matrix}sin^3x=cos^3x\\cos2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\cos2x=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=0\end{matrix}\right.\)
Giải PT
a) sin2 x + 2sinx - 3 = 0
b) 2cos x + cos 2x = 0
c) tanx + cotx + 2 = 0
d) sinx + cos2x + 1 = 0
e) tan x + cot 2x = 0
a) TH1: sinx = 1
--> x = pi/2 + k2pi (k nguyên)
TH2: sinx = -3 (loại)
b) 2cosx + cos2x = 0
<=> 2cosx + 2cos^2(x) - 1 = 0
TH1: cosx = (-1 + sqrt(3))/2
TH2: cosx = (-1 - sqrt(3))/2 (loại)
c) ĐKXĐ: x # kpi
Pt <=> tanx + 1/tanx + 2 = 0
--> tanx = -1
--> x = -pi/4 + kpi (k nguyên)
1)giải pt a)√2 cos2x-1=0
b) sinx =cos3x
c) cos (x+π/3) +sin(3x+π/4)=0
d)tan 2x = cot (x+π/4)
e) sin x = √3 cos x
f) tan^2(π/3-2x)-3=0
a: \(\Leftrightarrow cos2x=\dfrac{1}{\sqrt{2}}\)
=>2x=pi/4+k2pi hoặc 2x=-pi/4+k2pi
=>x=pi/8+kpi hoặc x=-pi/8+kpi
b: \(\Leftrightarrow sinx=sin\left(\dfrac{pi}{2}-3x\right)\)
=>x=pi/2-3x+k2pi hoặ x=pi/2+3x+k2pi
=>4x=pi/2+k2pi hoặc -2x=pi/2+k2pi
=>x=pi/8+kpi/2 hoặc x=-pi/4-kpi
d: \(\Leftrightarrow cos\left(x+\dfrac{pi}{3}\right)=-sin\left(3x+\dfrac{pi}{4}\right)\)
\(\Leftrightarrow cos\left(x+\dfrac{pi}{3}\right)=sin\left(-3x-\dfrac{pi}{4}\right)\)
\(\Leftrightarrow cos\left(x+\dfrac{pi}{3}\right)=cos\left(3x+\dfrac{3}{4}pi\right)\)
=>3x+3/4pi=x+pi/3+k2pi hoặc 3x+3/4pi=-x-pi/3+k2pi
=>2x=-5/12pi+k2pi hoặc 4x=-13/12pi+k2pi
=>x=-5/24pi+kpi hoặc x=-13/48pi+kpi/2
e: \(\Leftrightarrow sinx-\sqrt{3}\cdot cosx=0\)
\(\Leftrightarrow sin\left(x-\dfrac{pi}{3}\right)=0\)
=>x-pi/3=kpi
=>x=kpi+pi/3
Giải các phương trình sau:
a, \(\sqrt{2}\) sin \(\left(2x+\frac{\pi}{4}\right)\)=3sinx+cosx+2
b, 1+sinx+cosx+sin2x+cos2x=0
c, (2cosx-1)(2sinx+cosx)=sin2x-sinx
d, cos3x+cos2x-cosx-1=0
a.
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(sinx+cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=-1\\2cosx-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\\cosx=\frac{3}{2}\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
\(\Leftrightarrow1+sinx+cosx+2sinx.cosx+2cos^2x-1=0\)
\(\Leftrightarrow sinx\left(2cosx+1\right)+cosx\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
c.
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)=2sinx.cosx-sinx\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)-sinx\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx-sinx\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx+cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
Giải các phương trình lượng giác sau:
1) a/ \(cos\left(10x+12\right)+4\sqrt{2}sin\left(5x+6\right)-4=0\)
b/ \(cos\left(4x+2\right)+3sin\left(2x+1\right)=2\)
2) a/ \(cos2x+sin^2x+2cosx+1=0\)
b/ \(4sin^22x-8cos^2x+ 3=0\)
c/ \(4cos2x+4sin^2x+4sinx=1\)
3) a/ \(tanx+cotx=2\)
b/ \(2tanx-2cotx=3\)
4) a/ \(2sin2x+8tanx=9\sqrt{3}\)
b/ \(2cos2x+tan^2x=5\)
5) a/ \(\left(3+cotx\right)^2=5\left(3+cotx\right)\)
b/ \(4\left(sin^2x+\dfrac{1}{sin^2x}\right)-4\left(sinx+\dfrac{1}{sinx}\right)=7\)
1a.
Đặt \(5x+6=u\)
\(cos2u+4\sqrt{2}sinu-4=0\)
\(\Leftrightarrow1-2sin^2u+4\sqrt{2}sinu-4=0\)
\(\Leftrightarrow2sin^2u-4\sqrt{2}sinu+3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinu=\dfrac{3\sqrt{2}}{2}>1\left(loại\right)\\sinu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow sin\left(5x+6\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+6=\dfrac{\pi}{4}+k2\pi\\5x+6=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}+\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=-\dfrac{6}{5}+\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)
1b.
Đặt \(2x+1=u\)
\(cos2u+3sinu=2\)
\(\Leftrightarrow1-2sin^2u+3sinu=2\)
\(\Leftrightarrow2sin^2u-3sinu+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinu=1\\sinu=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(2x+1\right)=1\\sin\left(2x+1\right)=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{\pi}{2}+k2\pi\\2x+1=\dfrac{\pi}{6}+k2\pi\\2x+1=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}+\dfrac{\pi}{4}+k\pi\\x=-\dfrac{1}{2}+\dfrac{\pi}{12}+k\pi\\x=-\dfrac{1}{2}+\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)
2a.
\(cos^2x-sin^2x+sin^2x+2cosx+1=0\)
\(\Leftrightarrow cos^2x+2cosx+1=0\)
\(\Leftrightarrow\left(cosx+1\right)^2=0\)
\(\Leftrightarrow cosx=-1\)
\(\Leftrightarrow x=\pi+k2\pi\)
Giải các phương trình sau:
a, \(\dfrac{Sin^2x+Sinx}{Sinx-1}=-2\)
b,\(\dfrac{Cos2x+Sinx}{Sinx-1}+1=0\)
a)Đk:\(sinx\ne1\)
Pt\(\Leftrightarrow sin^2x+sinx=-2\left(sinx-1\right)\)
\(\Leftrightarrow sin^2x+3sinx-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{-3+\sqrt{17}}{2}\left(tm\right)\\sinx=\dfrac{-3-\sqrt{17}}{2}\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arcc.sin\left(\dfrac{-3+\sqrt{17}}{2}\right)+k2\pi\\x=\pi-arc.sin\left(\dfrac{-3+\sqrt{17}}{2}\right)+k2\pi\end{matrix}\right.\)(\(k\in Z\))
b)Đk:\(sinx\ne1\)
Pt \(\Leftrightarrow\dfrac{1-2sin^2x+sinx}{sinx-1}+1=0\)
\(\Leftrightarrow\dfrac{-\left(sinx-1\right)\left(2sinx+1\right)}{sinx-1}+1=0\)
\(\Leftrightarrow-\left(2sinx+1\right)+1=0\)
\(\Leftrightarrow sinx=0\) (tm)
\(\Leftrightarrow x=k\pi,k\in Z\)
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