• Cos(3 - 2x) + cosx =0
• Cos(3 - 2x) - sinx =0
• Cos(3 + 2x) + sinx =0
• Sin4x - √3sin2x =0
M.n giúp mình với ạ 💓
sin^3 x +cos^3 x -3sinx cosx+1=0
3 cosx -3sin2x= √3(cos2x+sinx)
4sin^3x +3sin^2x cosx -sinx-cos^3x=0
√3sin4x-cos4x=sinx- √3cosx
m.n giúp mk chứng minh với ạ
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.
\(\Leftrightarrow\left(1+sinx\right)\left(cosx-sinx\right)=1-sin^2x\)
\(\Leftrightarrow\left(1+sinx\right)\left(cosx-sinx\right)-\left(1+sinx\right)\left(1-sinx\right)=0\)
\(\Leftrightarrow\left(1+sinx\right)\left(cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{2}+k2\pi\\x=k2\pi\end{matrix}\right.\)
giải phương trình
a. cosx - \(\sqrt{3}\)sinx = \(\sqrt{2}\)
b. 5\(sin^2x\) + sinxcosx - 6\(cos^2x\)=0
a/ \(\dfrac{1}{2}\cos x-\dfrac{\sqrt{3}}{2}\sin x=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sin\left(\dfrac{\pi}{6}-x\right)=\dfrac{\sqrt{2}}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{6}-x=\dfrac{\pi}{4}+k2\pi\\\dfrac{\pi}{6}-x=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k2\pi\\x=-\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\)
b/ \(\cos x=0\) ko la nghiem cua pt
\(\cos x\ne0\Rightarrow pt\Leftrightarrow5\tan^2x+\tan x-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\tan x=1\\\tan x=-\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow...\)
1:\(\left(sin\dfrac{x}{2}+cos\dfrac{x}{2}\right)^2+\sqrt{3}cosx=2\)
2: \(cos^2x-\sqrt{3}sin2x=1+sin^2x\)
3: \(4\left(sin^4x+cos^4x\right)+\sqrt{3}sin4x=2\)
4:\(cos5x-2sin3xcos2x-sinx=0\)
\(1.\left(sinx+cosx\right)^3+sinxcosx-1=0\)
\(2.\left(sinx+cosx\right)^4-3sin2x-1=0\)
\(3.sin^3x+cos^3x+2\left(sinx+cosx\right)-3sin2x=0\)
\(4.\left(sinx-cosx\right)^3=1+sinxcosx\)
5.\(sinx+cosx+2+tanx+cotx+\frac{1}{sinx}+\frac{1}{cosx}=0\)
1.
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
Pt trở thành:
\(t^3+\frac{t^2-1}{2}-1=0\)
\(\Leftrightarrow2t^3+t^2-3=0\)
\(\Leftrightarrow\left(t-1\right)\left(2t^2+3t+3\right)=0\)
\(\Leftrightarrow t=1\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sin2x=2sinx.cosx=t^2-1\end{matrix}\right.\)
Pt trở thành:
\(t^4-3\left(t^2-1\right)-1=0\)
\(\Leftrightarrow t^4-3t^2+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t^2=1\\t^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}1+sin2x=1\\1+sin2x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sin2x=1\end{matrix}\right.\)
\(\Leftrightarrow...\)
3.
\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)+2\left(sinx+cosx\right)-6sinx.cosx=0\)
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
Pt trở thành:
\(t\left(1-\frac{t^2-1}{2}\right)+2t-3\left(t^2-1\right)=0\)
\(\Leftrightarrow-t^3-6t^2+7t+6=0\)
Nghiệm của pt bậc 3 này rất xấu, chắc bạn ghi ko đúng đề bài
Tìm GTLN và GTNN của hàm số : 1. y = sinx + 2cosx +1 / 2sinx + cosx + 3
2.y= 2sin^2sinx - 3 sinx cosx + cos^2 x
Giải phương trình : 1. 2sin^2 * 2x + sin7x -1 = sinx
2.cos 4x + 12 sin^2 x -1 = 0
a)căn 3 sin4x-cos4x-2cosx=0
b)cosx +căn 3 cos2x-căn 3 sinx-sin2x=0
c)cos 3x+sin2x=căn 3(sin3x+cos2x)
d)cosx +căn 3=3-3/cosx+căn 3 sinx+1
a/
\(\sqrt{3}sin4x-cos4x=2cosx\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin4x-\frac{1}{2}cos4x=cosx\)
\(\Leftrightarrow sin\left(4x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{6}=\frac{\pi}{2}-x+k2\pi\\4x-\frac{\pi}{6}=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{15}+\frac{k2\pi}{5}\\x=\frac{2\pi}{9}+\frac{k2\pi}{3}\end{matrix}\right.\)
b/
\(\Leftrightarrow cosx-\sqrt{3}sinx=sin2x-\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=sin\left(2x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=sin\left(\frac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{5\pi}{6}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow cos3x-\sqrt{3}sin3x=\sqrt{3}cos2x-sin2x\)
\(\Leftrightarrow\frac{1}{2}cos3x-\frac{\sqrt{3}}{2}sin3x=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow cos\left(3x+\frac{\pi}{3}\right)=cos\left(2x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{3}=2x+\frac{\pi}{6}+k2\pi\\3x+\frac{\pi}{3}=-2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=-\frac{\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
1) Cos^4x-sin^4x+5cosx+3=0
2) 3cos^2x+cos^2x×sinx=8(1+sinx)
1.
\(\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+5cosx+3=0\)
\(\Leftrightarrow2cos^2x-1+5cosx+3=0\)
\(\Leftrightarrow2cos^2x+5cosx+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\frac{1}{2}\\cosx=-2\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
\(3\left(1-sin^2x\right)+\left(1-sin^2x\right)sinx=8\left(1+sinx\right)\)
\(\Leftrightarrow\left(1+sinx\right)\left(3-3sinx\right)+\left(1+sinx\right)\left(sinx-sin^2x\right)=8\left(1+sinx\right)\)
\(\Leftrightarrow\left(1+sinx\right)\left(3-3sinx+sinx-sin^2x-8\right)=0\)
\(\Leftrightarrow\left(1+sinx\right)\left(-sin^2x-2sinx-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\-sin^2x-2sinx-5=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
1, 3sinx - 4cosx =1
2, \(\sqrt{3}\)sinx - cosx =1
3, \(\sqrt{3}\)cosx + sinx = -2
4, cos4x - sin4x = 1
5, \(\sqrt{3}\)cos4x + sin4x - 2cos3x = 0
6, cos2x= 3sin2x + 3
7, 3sin5x - 2cos5x = 3
\(\text{1) }3sinx-4cosx=1\\ \Leftrightarrow cos^2x+\left(\frac{4cosx+1}{3}\right)^2=1\\ \Leftrightarrow cosx=\frac{-4\pm6\sqrt{6}}{25}\\ \\ \Leftrightarrow x=arccos\left(\frac{-4\pm6\sqrt{6}}{25}\right)+k2\pi\)
\(2\text{) }\sqrt{3}sinx-cosx=1\\ \Leftrightarrow\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx=\frac{1}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sinx-sin\frac{\pi}{6}\cdot cosx=\frac{1}{2}\\ \Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=sin\frac{\pi}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=\frac{\pi}{6}+a2\pi\\x-\frac{\pi}{6}=\frac{5\pi}{6}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+a2\pi\\x=\pi+b2\pi\end{matrix}\right.\)
\(3\text{) }\sqrt{3}cosx+sinx=-2\\ \Leftrightarrow\frac{\sqrt{3}}{2}cosx+\frac{1}{2}sinx=-1\\ \Leftrightarrow sin\frac{\pi}{3}\cdot cosx+cos\frac{\pi}{3}\cdot sinx=-1\\ \Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=-1=sin\frac{3\pi}{2}\\ \\ \Leftrightarrow x+\frac{\pi}{3}=\frac{3\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{7\pi}{6}+k2\pi\)
\(4\text{) }cos4x-sin4x=1\\ \Leftrightarrow cos^24x+\left(cos4x-1\right)^2=1\\ \\ \Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+a\pi\\4x=b2\pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{a\pi}{4}\\x=\frac{b\pi}{2}\end{matrix}\right.\)
\(5\text{) }\sqrt{3}cos4x+sin4x-2cos3x=0\\ \Leftrightarrow\frac{\sqrt{3}}{2}cos4x+\frac{1}{2}sin4x=cos3x\\ \Leftrightarrow cos\frac{\pi}{3}\cdot cos4x+sin\frac{\pi}{3}\cdot sin4x=cos3x\\ \Leftrightarrow cos\left(4x-\frac{\pi}{3}\right)=cos3x\\ \Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{3}=3x+a2\pi\\4x-\frac{\pi}{3}=-3x+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+a2\pi\\x=\frac{\pi}{21}+\frac{b2\pi}{7}\end{matrix}\right.\\ \Leftrightarrow x=\frac{\pi}{21}+\frac{k2\pi}{7}\)
\(6\text{) }cos^2x=3sin2x+3\\ \Leftrightarrow\frac{cos2x+1}{2}=3sin2x+3\)
Giải tương tự vd 1 và 4
7) Giải tương tự vd 1 và 4