Giải phương trình lượng giác sau:
sinx+cosx=\(\frac{cos2x}{sin^2x}\)
x=?
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 lượng giác:
1.sin^2x + sin 2x = 3 cos^2x
2.sinx + cosx = 2√2 sinxcosx
1. \(\sin^2x+\sin2x=3\cos^2x\Leftrightarrow\sin^2x+2\sin x\cos x-3\cos^2x=0\Leftrightarrow4\sin^2x+2\sin x\cos x-3=0\)
Vì \(\cos x=0\) không phải là nghiệm của phương trình, nên chia 2 vế pt cho \(\cos x\), ta đc:
\(4\tan^2x+2\tan x-\frac{3}{\cos^2x}=0\Leftrightarrow4\tan^2x+2\tan x-3\left(1+\tan^2x\right)=0\Leftrightarrow\tan^2x+2\tan x-3=0\)
Suy ra: \(\begin{matrix}\tan x=1\\\tan x=-3\end{matrix}\) suy ra x.
b) \(\Leftrightarrow\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)=\sqrt{2}\sin2x\Leftrightarrow\sin\left(x+\frac{\pi}{4}\right)=\sin2x\Leftrightarrow\begin{cases}x+\frac{\pi}{4}=2x+k2\pi\\x+\frac{\pi}{4}=\pi-2x+k2\pi\end{cases}\)
\(\Leftrightarrow\begin{cases}x=\frac{\pi}{4}-k2\pi\\x=\frac{\pi}{4}+\frac{k2\pi}{3}\end{cases}\)
Vậy ....
Chỗ Viết các nghiệm: Sửa lại : dùng dấu ngoặc vuông thay cho ngoặc nhọn
Giải phương trình: Sin^3x + cos^3x - sinx - cosx = cos2x
\(sin^3x+cos^3x-sinx-cosx=cos2x\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sin^2x-sinx.cosx+cos^2x\right)-\left(sinx+cosx\right)-\left(cos^2x-sin^2x\right)\)\(=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)-\left(sinx+cosx\right)-\left(cosx+sinx\right)\left(cosx-sinx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sinx-cosx-sinx.cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\left(1\right)\\sinx-cosx-sinx.cosx=0\left(2\right)\end{matrix}\right.\)
TH1: (1)\(\Leftrightarrow\sqrt{2}.sin\left(x+\dfrac{\pi}{4}\right)=0\)\(\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\left(k\in Z\right)\)
TH2: Đặt \(t=sinx-cosx\) ;\(t\in\left(-2;2\right)\)
\(\Rightarrow\dfrac{t^2-1}{2}=-sinx.cosx\)
Pt (2)\(\Rightarrow t+\dfrac{t^2-1}{2}=0\)\(\Leftrightarrow t^2+2t-1=0\) \(\Leftrightarrow\left[{}\begin{matrix}t=-1+\sqrt{2}\left(tm\right)\\t=-1-\sqrt{2}\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow sinx-cosx=-1+\sqrt{2}\)\(\Leftrightarrow\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)=-\sqrt{2}+1\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{1-\sqrt{2}}{\sqrt{2}}\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+arc.cos\dfrac{1-\sqrt{2}}{2}+k2\pi\\x=\dfrac{-\pi}{4}-arc.cos\dfrac{1-\sqrt{2}}{2}+k2\pi\end{matrix}\right.\)(\(k\in\)\(Z\))
Vậy...
giải phương trình
1.\(sin^3x+2cosx-2+sin^2x=0\)
\(2.\frac{\sqrt{3}}{2}sin2x+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)
3.\(2sin2x-cos2x=7sinx+2cosx-4\)
4.\(2cos2x-8cosx+7=\frac{1}{cosx}\)
5.\(cos^8x+sin^8x=2\left(cos^{10}x+sin^{10}x\right)+\frac{5}{4}cos2x\)
6.\(1+sinx+cos3x=cosx+sin2x+cos2x\)
7.\(1+sinx+cosx+sin2x+cos2x=0\)
1.
\(\Leftrightarrow sin^2x\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cos^2x\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(1+cosx\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(sinx+cosx+sinx.cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\Leftrightarrow...\\sinx+cosx+sinx.cosx-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
Đặt \(sinx+cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow t+\frac{t^2-1}{2}-1=0\)
\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
2.
\(\Leftrightarrow\sqrt{3}sinx.cosx+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)
\(\Leftrightarrow cosx\left(\sqrt{3}sinx+\sqrt{2}cosx+\sqrt{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Leftrightarrow...\\\sqrt{3}sinx+\sqrt{2}cosx=-\sqrt{6}\left(1\right)\end{matrix}\right.\)
Xét (1):
Do \(\sqrt{3}^2+\sqrt{2}^2< \left(-\sqrt{6}\right)^2\) nên (1) vô nghiệm
3.
\(\Leftrightarrow4sinx.cosx-\left(1-2sin^2x\right)=7sinx+2cosx-4\)
\(\Leftrightarrow2cosx\left(2sinx-1\right)+2sin^2x-7sinx+3=0\)
\(\Leftrightarrow2cosx\left(2sinx-1\right)+\left(sinx-3\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2cosx+sinx-3\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\Leftrightarrow...\\2cosx+sinx=3\left(1\right)\end{matrix}\right.\)
Xét (1), do \(2^2+1^2< 3^2\) nên (1) vô nghiệm
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.
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)=2sinx.cosx-sinx\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)-sinx\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx-sinx\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx+cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
Giaỉ các phương trình lượng giác sau:
1. sin(sinx)=0
2. sin(cosx)=0
3. \(\sqrt{3}\sin-\cos x=2cos3x\)
4. \(\sin2x=sin\left(2x-\dfrac{\pi}{2}\right)\)
5. \(4\cos\left(3\pi-2x\right)=\sqrt{2}\)
3.
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=cos3x\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{2}-3x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{2}-3x+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)
câu 2 mình sửa lại đề bài một chút là: sin(cosx)=1 ạ
1.
\(sin\left(sinx\right)=0\)
\(\Leftrightarrow sinx=k\pi\) (1)
Do \(-1\le sinx\le1\Rightarrow-1\le k\pi\le1\)
\(\Rightarrow-\dfrac{1}{\pi}\le k\le\dfrac{1}{\pi}\Rightarrow k=0\) do \(k\in Z\)
Thế vào (1)
\(\Rightarrow sinx=0\Rightarrow x=n\pi\)
2.
\(sin\left(cosx\right)=1\Leftrightarrow cosx=\dfrac{\pi}{2}+k2\pi\)
Do \(-1\le cosx\le1\Rightarrow-1\le\dfrac{\pi}{2}+k2\pi\le1\)
\(\Rightarrow-\dfrac{1}{2\pi}-\dfrac{1}{4}\le k\le\dfrac{1}{2\pi}-\dfrac{1}{4}\)
\(\Rightarrow\) Không tồn tại k thỏa mãn
Pt vô nghiệm
HELPING NOW!!!
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 phương trình sau:
a, sinx+cosx+1+sin2x+cos2x=0
b, sinx(1+cos2x)+sin2x=1+cos2x
c, \(\frac{1}{sinx}+\frac{1}{sin\left(x-\frac{3\pi}{2}\right)}=4sin\left(\frac{7\pi}{4}-x\right)\)
d, sin4x+cos4x=\(\frac{7}{8}cot\left(x+\frac{\pi}{3}\right)cot\left(\frac{\pi}{6}-x\right)\)
@Nguyễn Việt Lâm giúp em với ạ
a.
\(sinx+cosx+\left(sinx+cosx\right)^2+cos^2x-sin^2x=0\)
\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1+2cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\1+2cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
b.
\(sinx\left(1+2cos^2x-1\right)+2sinx.cosx=1+2cos^2x-1\)
\(\Leftrightarrow cos^2x.sinx+sinx.cosx-cos^2x=0\)
\(\Leftrightarrow cosx\left(sinx.cosx+sinx-cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\sinx.cosx+sinx-cosx=0\left(1\right)\end{matrix}\right.\)
Xét (1), đặt \(sinx-cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{1-t^2}{2}\end{matrix}\right.\)
\(\Rightarrow\frac{1-t^2}{2}+t=0\)
\(\Leftrightarrow-t^2+2t+1=0\Rightarrow\left[{}\begin{matrix}t=1-\sqrt{2}\\t=1+\sqrt{2}>\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1-\sqrt{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=\frac{1-\sqrt{2}}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+arcsin\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)+k2\pi\\x=\frac{5\pi}{4}-arcsin\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
giải phương trình sau:
\(\dfrac{2sin^2x+cos4x-cos2x}{\left(sinx-cosx\right)sin2x}\)=0
ĐK: \(x\ne\dfrac{\pi}{4}+k\pi;x\ne\dfrac{k\pi}{2}\)
\(\dfrac{2sin^2x+cos4x-cos2x}{\left(sinx-cosx\right)sin2x}=0\)
\(\Leftrightarrow2sin^2x+cos4x-cos2x=0\)
\(\Leftrightarrow2sin^2x-1+cos4x-cos2x+1=0\)
\(\Leftrightarrow2cos^22x-2cos2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\2x=k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=k\pi\end{matrix}\right.\)
Đối chiếu điều kiện ta được \(x=-\dfrac{\pi}{4}+k\pi\)