giải phương trình:
1) \(2\sqrt{2}cos^3x\left(x-\frac{\pi}{4}\right)-3cosx-sinx=0\)
2) \(tanx.sin^2x-2sin^2x=3\left(cos2x+sinxcosx\right)\)
3) \(2sin^3x=cosx\)
4) \(6sinx-2cos^3x=\frac{5sin4xcosx}{2cos2x}\)
giải phương trình đối với sin x và cosx
1) 3sinx-4cosx=5
2) \(\sqrt{3}cos2x+sin2x+2sin\left(2x-\frac{\pi}{6}\right)=2\sqrt{2}\)
3) \(cosx+\sqrt{3}sinx+2cos\left(2x+\frac{\pi}{3}\right)=0\)
4) \(2cos\left(2x+\frac{\pi}{6}\right)+4sinxcosx-1=0\)
5) \(\sqrt{3}cos5x-2sin3x.cos2x-sinx=0\)
giải các pt
a) \(\left(1+tanx\right)sin^2x=3sinx\left(cosx-sinx\right)+3\)
b) \(6sinx-2cos^3x=\frac{5sin4x.sinx}{2cos2x}\)
c) \(cos^3x=2sinx.sin\left(\frac{\pi}{3}-x\right).sin\left(x+\frac{\pi}{3}\right)\)
d) \(cos2x\left(sinx+cosx\right)-4cos^3x\left(1+sin2x\right)=0\)
a.
ĐKXĐ: \(cosx\ne0\)
Chia 2 vế cho \(cos^2x\) ta được:
\(\left(1+tanx\right).tan^2x=3tanx\left(1-tanx\right)+\frac{3}{cos^2x}\)
\(\Leftrightarrow tan^2x\left(tanx+1\right)=3tanx-3tan^2x+3+3tan^2x\)
\(\Leftrightarrow tan^2x\left(tanx+1\right)-3\left(tanx+1\right)=0\)
\(\Leftrightarrow\left(tan^2x-3\right)\left(tanx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=\sqrt{3}\\tanx=-\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow cos^3x=sinx\left(cos\frac{2\pi}{3}+cos2x\right)\)
\(\Leftrightarrow cos^3x=sinx\left(cos2x-\frac{1}{2}\right)\)
\(\Leftrightarrow cos^3x=2sinx\left(1-2sin^2x-\frac{1}{2}\right)\)
\(\Leftrightarrow cos^3x=sinx\left(\frac{1}{2}-2sin^2x\right)\)
\(\Leftrightarrow2cos^3x=sinx-4sin^3x\)
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)
\(\Leftrightarrow2=tanx\left(1+tan^2x\right)-4tan^3x\)
\(\Leftrightarrow3tan^3x-tanx+2=0\)
\(\Leftrightarrow\left(tanx+1\right)\left(3tan^2x-3tanx+2\right)=0\)
\(\Leftrightarrow tanx=-1\Rightarrow x=-\frac{\pi}{4}+k\pi\)
d/
\(\Leftrightarrow\left(cos^2x-sin^2x\right)\left(sinx+cosx\right)-4cos^3x\left(sin^2x+cos^2x+2sinx.cosx\right)=0\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(sinx+cosx\right)^2-4cos^3x\left(sinx+cosx\right)^2=0\)
\(\Leftrightarrow\left(cosx-sinx-4cos^3x\right)\left(sinx+cosx\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\left(1\right)\\cosx-sinx-4cos^3x=0\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=0\Leftrightarrow x+\frac{\pi}{4}=k\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
Xét \(\left(2\right)\), nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)
\(\Leftrightarrow\frac{1}{cos^2x}-tanx.\frac{1}{cos^2x}-4=0\)
\(\Leftrightarrow1+tan^2x-tanx\left(1+tan^2x\right)-4=0\)
\(\Leftrightarrow-tan^3x+tan^2x-tanx-3=0\)
\(\Leftrightarrow\left(tanx+1\right)\left(tan^2x-2tanx+3\right)=0\)
\(\Leftrightarrow tanx=-1\Rightarrow x=-\frac{\pi}{4}+k\pi\)
giải các phương trình sau:
1) \(\sqrt{3}sin^2x+\left(1-\sqrt{3}\right)sinxcosx-cos^2x+1-\sqrt{3}=0\)
2) \(9sin^2x-30sinxcosx+25cos^2x=25\)
3) \(sin2x-2sin^2x=2cos2x\)
4) \(sin^3x-cos^3x=sinx+cosx\)
5)\(4\left(sin^3x+cos^3x\right)=sinx+cosx\)
mik lm biếng quá mik chỉ nói cách làm thôi nha bạn
1) chia hai vế cho cos^2(x) \(\sqrt{3}tan^2x+\left(1-\sqrt{3}\right)tanx-1+\left(1-\sqrt{3}\right)\left(1+tan^2x\right)=0\)
đặt t = tanx rr giải thôi =D ( máy 570 thì mode5 3 còn máy 580 thì mode 9 2 2) :)))
2) cx làm cách tương tự chia 2 vế cho cos^2x
3) giữ vế trái bung vế phải ra
\(sin2x-2sin^2x=2-4sin^22x\)
đặt t = sin2x (-1=<t=<1)
4) đẩy sinx cosx qua trái hết
\(sinx\left(sin^2-1\right)-cosx\left(cos^2x+1\right)=0\)
\(sinx\left(-cos^2x\right)-cos\left(cos^2x+1\right)=0\)
\(-cos\left(sinxcosx+cos^2x+1\right)=0\)
cái vế đầu cosx=0 bn bik giả rr mà dễ ẹc à còn vế sau thì chia cho cos^2(x) như mấy bài trên rr sau đó đặt t = tanx rr bấm máy là ra thui :))
5)bung cái hằng đẳng thức ra sau đó đặt t=sinx+cosx (t thuộc [-căn(2) ; căn(2)]
khi đó ta có sinxcosx=1/2 sin2x= 1/2t^2 - 1/2
làm đi là ra à
Giải các pt sau
a, \(\dfrac{1}{sinx}+\dfrac{1}{cosx}=4sin\left(x+\dfrac{\pi}{4}\right)\)
b, \(2sin\left(2x-\dfrac{\pi}{6}\right)+4sinx+1=0\)
c, \(cos2x+\sqrt{3}sinx+\sqrt{3}sin2x-cosx=2\)
d, \(4sin^2\dfrac{x}{2}-\sqrt{3}cos2x=1+cos^2\left(x-\dfrac{3\pi}{4}\right)\)
a)\(\dfrac{2sin^2\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)+\sqrt{3}cos^3x\left(1-3tan^2x\right)}{2sinx-1}=-1\)
b)\(\dfrac{2sin2x-cos2x-7sinx+4+\sqrt{3}}{2cosx+\sqrt{3}}=1\)
c)\(\dfrac{\left(1+sinx+cos2x\right)sin\left(x+\dfrac{\pi}{4}\right)}{1+tanx}=\dfrac{1}{\sqrt{2}}cosx\)
d)\(\left(\sqrt{3}sin2x+1\right)\left(2sinx-1\right)+sin3x-cos2x-sinx=0\)
a, ĐK: \(x\ne\dfrac{5\pi}{6}+k2\pi;x\ne\dfrac{\pi}{6}+k2\pi\)
\(\dfrac{2sin^2\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)+\sqrt{3}cos^3x\left(1-3tan^2x\right)}{2sinx-1}=-1\)
\(\Leftrightarrow2sin^2\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)+\sqrt{3}cos^3x\left(1-3tan^2x\right)=1-2sinx\)
\(\Leftrightarrow-cos\left(3x-\dfrac{\pi}{2}\right)+\sqrt{3}cos^3x.\dfrac{cos^2x-3sin^2x}{cos^2x}=-2sinx\)
\(\Leftrightarrow-sin3x+\sqrt{3}cosx.\left(cos^2x-3sin^2x\right)=-2sinx\)
\(\Leftrightarrow-sin3x+\sqrt{3}cosx.\left(4cos^2x-3\right)=-2sinx\)
\(\Leftrightarrow-sin3x+\sqrt{3}cos3x=-2sinx\)
\(\Leftrightarrow\dfrac{1}{2}sin3x-\dfrac{\sqrt{3}}{2}cos3x-sinx=0\)
\(\Leftrightarrow sin\left(3x-\dfrac{\pi}{3}\right)-sinx=0\)
\(\Leftrightarrow2cos\left(2x-\dfrac{\pi}{6}\right)sin\left(x-\dfrac{\pi}{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(2x-\dfrac{\pi}{6}\right)=0\\sin\left(x-\dfrac{\pi}{6}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi\\x-\dfrac{\pi}{6}=k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
Đối chiếu điều kiện ta được:
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\\x=\dfrac{7\pi}{6}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
giải phương trình:
a, \(tanx.sin^2x-2sin^2x=3\left(cos2x+sinxcosx\right)\)
b, \(5sinx-2=3\left(1-sinx\right)tan^2x\)
c,\(\frac{cos2x+3cot2x+4sinx}{cot2x-cos2x}=2\)
d, \(\frac{4sin^2x+6sin^2x-3cos2x-9}{cosx}=0\)
Tìm giá trị lớn nhất và giá trị nhỏ nhất của các hàm số sau:
1,\(y=5-3cosx\)
2,\(y=3cos^2x-2cosx+2\)
3,\(y=cos^2x+2cos2x\)
4,\(y=\sqrt{5-2sin^2x.cos^2x}\)
5,\(y=cos2x-cos\left(2x-\dfrac{\pi}{3}\right)\)
6,\(y=\sqrt{3}sinx-cosx-2\)
7,\(y=2cos^2x-sin2x+5\)
8,\(y=2sin^2x-sin2x+10\)
9,\(y=sin^6x+cos^6x\)
giải các pt
a) \(cos^2\left(\frac{\pi}{3}+x\right)+4cos\left(\frac{\pi}{6}-x\right)=4\)
b) \(5cos\left(2x+\frac{\pi}{3}\right)=4sin\left(\frac{5\pi}{6}-x\right)-9\)
c) \(2sin^2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
d) \(2sin^2x+\sqrt{3}sin2x+4=4\left(\sqrt{3}sinx+cosx\right)\)
a/
Đặt \(x+\frac{\pi}{3}=a\Rightarrow x=a-\frac{\pi}{3}\)
Pt trở thành:
\(cos^2a+4cos\left(\frac{\pi}{6}-a+\frac{\pi}{3}\right)=4\)
\(\Leftrightarrow cos^2a+4cos\left(\frac{\pi}{2}-a\right)-4=0\)
\(\Leftrightarrow cos^2a+4sina-4=0\)
\(\Leftrightarrow1-sin^2a+4sina-4=0\)
\(\Leftrightarrow-sin^2a+4sina-3=0\)
\(\Rightarrow\left[{}\begin{matrix}sina=1\\sina=3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow sin\left(x+\frac{\pi}{3}\right)=1\)
\(\Rightarrow x+\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\frac{\pi}{6}+k2\pi\)
b/
Đặt \(x+\frac{\pi}{6}=a\Rightarrow x=a-\frac{\pi}{6}\)
Pt trở thành:
\(5cos2a=4sin\left(\frac{5\pi}{6}-a+\frac{\pi}{6}\right)-9\)
\(\Leftrightarrow5cos2x=4sin\left(\pi-a\right)-9\)
\(\Leftrightarrow5\left(1-2sin^2a\right)=4sina-9\)
\(\Leftrightarrow10sin^2a+4sina-14=0\)
\(\Rightarrow\left[{}\begin{matrix}sina=1\\sina=-\frac{7}{5}< -1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow sin\left(x+\frac{\pi}{6}\right)=1\)
\(\Rightarrow x+\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\frac{\pi}{3}+k2\pi\)
c/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=...\)
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
⇔ \(\left[{}\begin{matrix}sin^3x=cos^3x\\cos2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\cos2x=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=0\end{matrix}\right.\)