3sin^2x + 4sin2x +(8√3 -9) *cos^2x=0
sin^2 + sin2x - 2cos^2x =1/2
(sinx +1) *( 2cos 2x - 2) =0
giải hộ e bài này vs ạ
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 các phương trình sau:
a/ Cos(x-pi/3)-sin(x-pi/3)=1
b/ Căn 3 sin2x + 2cos^2x = 2 sinx +1
Giúp mk với ạ
a, \(cos\left(x-\dfrac{\pi}{3}\right)-sin\left(x-\dfrac{\pi}{3}\right)=1\)
\(\Leftrightarrow\sqrt{2}cos\left(x-\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow cos\left(x-\dfrac{7\pi}{12}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow x-\dfrac{7\pi}{12}=\pm\dfrac{\pi}{4}+k2\pi\)
...
b, \(\sqrt{3}sin2x+2cos^2x=2sinx+1\)
\(\Leftrightarrow\sqrt{3}sin2x+2cos^2x-1=2sinx\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin2x+\dfrac{1}{2}cos2x=sinx\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)=sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=x+k2\pi\\2x+\dfrac{\pi}{6}=\pi-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
a)\(sinx+cosx=\dfrac{1}{cosx}\)
b)\(4sin2x-3sin\left(2x-\dfrac{\pi}{2}\right)=5\)
c)\(sin2x+sin^2x=\dfrac{1}{2}\)
Giải hộ em 3 pt trên với! Em cảm ơn.
a.
ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)
Chia 2 vế cho cosx:
\(tanx+1=\dfrac{1}{cos^2x}\)
\(\Rightarrow tanx+1=1+tan^2x\)
\(\Rightarrow\left[{}\begin{matrix}tanx=0\\tanx=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)
c.
\(\Leftrightarrow2sin2x+2sin^2x=1\)
\(\Leftrightarrow2sin2x=1-2sin^2x\)
\(\Leftrightarrow2sin2x=cos2x\)
\(\Rightarrow tan2x=\dfrac{1}{2}\)
\(\Rightarrow2x=arctan\left(\dfrac{1}{2}\right)+k\pi\)
\(\Rightarrow x=\dfrac{1}{2}arctan\left(\dfrac{1}{2}\right)+\dfrac{k\pi}{2}\)
b.
\(\Leftrightarrow4sin2x+3sin\left(\dfrac{\pi}{2}-2x\right)=5\)
\(\Leftrightarrow4sin2x+3cos2x=5\)
\(\Leftrightarrow\dfrac{4}{5}sin2x+\dfrac{3}{5}cos2x=1\)
Đặt \(\dfrac{4}{5}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow\dfrac{3}{5}=sina\)
\(\Rightarrow sin2x.cosa+cos2x.sina=1\)
\(\Rightarrow sin\left(2x+a\right)=1\)
\(\Rightarrow2x+a=\dfrac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\dfrac{a}{2}+\dfrac{\pi}{4}+k\pi\)
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)
giải các pt sau
1. \(2sin^2x+5sĩnx.cosx+cos^2x=3\)
2 \(sin^2x-sin2x+2cos^2x=0\)
Giải phương trình
a) \(sin2x+\sqrt{2}sinx.sin2x=0\)
b) \(4sinx.cosx.cos2x-cos\frac{5x}{2}.sin\frac{3x}{2}=0\)
c) \(4sin3x+cosx-cos5x=0\)
d) \(2cos^2\left(x-\frac{\pi}{4}\right)+sin2x=0\)
e) \(sin\left(\frac{3\pi}{2}-sinx\right)=1\)
f) \(cos^2x-sin^2x+sin4x=0\)
a/
\(\Leftrightarrow sin2x\left(1+\sqrt{2}sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\1+\sqrt{2}sinx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sinx=-\frac{\sqrt{2}}{2}=sin\left(-\frac{\pi}{4}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\x=-\frac{\pi}{4}+k2\pi\\x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=-\frac{\pi}{4}+k2\pi\\x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow2sin2x.cos2x-\frac{1}{2}sin4x+\frac{1}{2}sinx=0\)
\(\Leftrightarrow sin4x-\frac{1}{2}sin4x+\frac{1}{2}sinx=0\)
\(\Leftrightarrow sin4x=-sinx=sin\left(-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}4x=-x+k2\pi\\4x=\pi+x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{5}\\x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{matrix}\right.\)
c/
\(\Leftrightarrow4sin3x+2sin3x.sin2x=0\)
\(\Leftrightarrow sin3x\left(2+sin2x\right)=0\)
\(\Leftrightarrow sin3x=0\) ( do \(2+sin2x>0;\forall x\))
\(\Leftrightarrow3x=k\pi\)
\(\Rightarrow x=\frac{k\pi}{3}\)
d/
\(2cos^2\left(x-\frac{\pi}{4}\right)+sin2x=0\)
\(\Leftrightarrow1+cos\left(2x-\frac{\pi}{2}\right)+sin2x=0\)
\(\Leftrightarrow1+sin2x+sin2x=0\)
\(\Leftrightarrow sin2x=-\frac{1}{2}=sin\left(-\frac{\pi}{6}\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x=-\frac{\pi}{6}+k2\pi\\2x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)
e/
\(sin\left(\frac{3\pi}{2}-sinx\right)=1\)
\(\Leftrightarrow\frac{3\pi}{2}-sinx=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow sinx=\pi+k2\pi\)
Mà \(-1\le sinx\le1\Rightarrow-1\le\pi+k2\pi\le1\)
\(\Rightarrow\) Không tồn tại k nguyên thỏa mãn
Pt đã cho vô nghiệm
f/
\(cos^2x-sin^2x+sin4x=0\)
\(\Leftrightarrow cos2x+2sin2x.cos2x=0\)
\(\Leftrightarrow cos2x\left(1+2sin2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin2x=-\frac{1}{2}=sin\left(-\frac{\pi}{6}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\2x=-\frac{\pi}{6}+k2\pi\\2x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)
1)\(sin^23x.cos2x+sin^2x=0\)
2)
\(cos^23x+cos^22x=sin^2x\)
3)
\(\frac{1}{4}+cos^2\frac{x}{3}=\frac{1}{2}sin^2\frac{x}{2}\)
4)
\(sin^23x-sin^22x-sin^2x=0\)
5)
\(2cos^2x=3sin^25x+2\)
6) 3cosx+2cos2x-cos3x=2sinxsin2x-1
7) \(sinx+cosx=\sqrt{2}\left(2-sin^32x\right)\)
1.
\(\Leftrightarrow\left(1-cos6x\right)cos2x+1-cos2x=0\)
\(\Leftrightarrow cos2x-cos2x.cos6x+1-cos2x=0\)
\(\Leftrightarrow\frac{1}{2}\left(cos8x-cos4x\right)-1=0\)
\(\Leftrightarrow2cos^24x-cos4x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=-1\\cos4x=\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow4x=\pi+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
2.
\(\Leftrightarrow1+cos6x+2cos^22x=1-cos2x\)
\(\Leftrightarrow cos6x+cos2x+2cos^22x=0\)
\(\Leftrightarrow cos4x.cos2x+cos^22x=0\)
\(\Leftrightarrow cos2x\left(cos4x+cos2x\right)=0\)
\(\Leftrightarrow cos2x\left(2cos^22x+cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=-1\\cos2x=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{2}+k\pi\\x=\pm\frac{\pi}{6}+k\pi\end{matrix}\right.\)
3.
Đặt \(\frac{x}{6}=t\Rightarrow\frac{1}{4}+cos^22t=\frac{1}{2}sin^23t\)
\(\Leftrightarrow1+4cos^22t=1-cos6t\)
\(\Leftrightarrow cos6t+4cos^22t=0\)
\(\Leftrightarrow4cos^32t+4cos^22t-3cos2t=0\)
\(\Leftrightarrow cos2t\left(4cos^22t+4cos2t-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2t=0\\cos2t=\frac{1}{2}\\cos2t=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{\pi}{4}+\frac{k\pi}{2}\\t=\pm\frac{\pi}{6}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{3}=\frac{\pi}{4}+\frac{k\pi}{2}\\\frac{x}{3}=\frac{\pi}{6}+k\pi\\\frac{x}{3}=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow x=...\)
Giải pt:
a, \(sin2x+2cos^2x=2\)
b, \(2sin^2x+sinx.cosx-cos^2x=0\)
a: =>sin2x+2*(1-cos2x)/2=2
=>sin2x-cos2x=1
=>căn 2*sin(2x-pi/4)=1
=>2x-pi/4=pi/4+k2pi hoặc 2x-pi/4=3/4pi+k2pi
=>x=pi/4+kpi hoặc x=pi/2+kpi
b: =>2*(1+cos2x)/2+1/2*sin2x-1/2(1-cos2x)=0
=>1+cos2x+1/2*sin2x-1/2+1/2cos2x=0
=>1/2*sin2x+3/2*cos2x=-1/2
=>sin(2x+a)=-cos(a)=cos(pi-a)
=>sin(2x+a)=sin(-pi/2+a)
=>2x+a=-pi/2+a+k2pi hoặc 2x+a=3/2pi-a+k2pi
=>x=-pi/4+kpi hoặc x=3/4pi-a+kpi
Giải phương trình sau
1.\(cos2x-\sqrt{3}sin2x=\sqrt{2}\)
2.\(4sin^2\frac{x}{2}-3\sqrt{3}sinx-2cos^2\frac{x}{2}=4\)
3. \(2\left(sinx+cosx\right)=4sinxcosx+1\)
4. \(cosx-sinx-2sin2x-1=0\)
\(5.\sqrt{3}sin2x+cos2x=2sinx\)
6. \(9sin^2x-5cos^2x-5sinx+4=0\)
7.\(cos^2x-\sqrt{3}sin2x=1+sinx\)
8.\(\frac{3}{cos^2x}=3+2tan^2x\)
1.
\(\frac{1}{2}cos2x-\frac{\sqrt{3}}{2}sin2x=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{3}=\frac{\pi}{4}+k2\pi\\2x+\frac{\pi}{3}=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{24}+k\pi\\x=-\frac{7\pi}{24}+k\pi\end{matrix}\right.\)
2.
\(2\left(1-cosx\right)-3\sqrt{3}sinx-\left(1+cosx\right)=4\)
\(\Leftrightarrow cosx+\sqrt{3}sinx=-1\)
\(\Leftrightarrow\frac{1}{2}cosx+\frac{\sqrt{3}}{2}sinx=-\frac{1}{2}\)
\(\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=-\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{2\pi}{3}+k2\pi\\x-\frac{\pi}{3}=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow...\)
3.
\(4sinx.cosx-2sinx+1-2cosx=0\)
\(\Leftrightarrow2sinx\left(2cosx-1\right)-\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
4.
\(cosx-sinx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\-4sinx.cosx=2t^2-2\end{matrix}\right.\)
Pt trở thành: \(t+2t^2-2-1=0\Leftrightarrow2t^2+t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-\frac{3}{2}< -\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}cos\left(x+\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\\x+\frac{\pi}{4}=-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow...\)
5.
\(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x=sinx\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=x+k2\pi\\2x+\frac{\pi}{6}=\pi-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
6.
\(9sin^2x-5\left(1-sin^2x\right)-5sinx+4=0\)
\(\Leftrightarrow14sin^2x-5sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=-\frac{1}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=arcsin\left(-\frac{1}{7}\right)+k2\pi\\x=\pi-arcsin\left(-\frac{1}{7}\right)+k2\pi\end{matrix}\right.\)