\(\cos5x+2\sin x\cos x+2\sin3x\sin2x=0\)
Giai các pt sau
1. \(\sqrt{3}\cos5x-2\sin3x.\cos2x-\sin x=0\)
4. \(\sin3x+\cos3x-\sin x+\cos x=\sqrt{2}\cos2x\)
6. \(\sin x+\cos x.\sin2x+\sqrt{3}\cos3x=2\left(\cos4x+\sin x^3\right)\)
sinx - sin3x + sin5x =0
sin2x + sin22x = sin23x
cos3x - cos5x = sinx
sin3x + sin5x + sin7x = 0
sinx + sin2x + sin3x - cosx - cos2x - cos3x = 0
Giải các phương trình sau :
a) \(\sin x+2\sin3x=-\sin5x\)
b) \(\cos5x\cos x=\cos4x\)
c) \(\sin x\sin2x\sin3x=\dfrac{1}{4}\sin4x\)
d) \(\sin^4x+\cos^4x=-\dfrac{1}{2}\cos^22x\)
Giải PT
a1) \(\dfrac{\left(1-2\sin x\right)\cos x}{\left(1+2\sin x\right)\left(1-\sin x\right)}=\sqrt{3}\)
a2) \(2\sin17x+\sqrt{3}\cos5x+\sin5x=0\)
a3) \(\)\(\cos7x-\sin5x=\sqrt{3}\left(\cos5x-\sin7x\right)\)
a4) \(\sqrt{3}\cos5x-2\sin3x\cos2x-\sin x=0\)
a5) \(\tan x+\cot x=2\left(\sin2x+\cos2x\right)\)
Giải các PT sau
1. \(\cos^2\left(x-30^{\cdot}\right)-\sin^2\left(x-30^{\cdot}\right)=\sin\left(x+60^{\cdot}\right)\)
2. \(\sin^22x+\cos^23x=1\)
3. \(\sin x+\sin2x+\sin3x+\sin4x=0\)
4. \(\sin^2x+\sin^22x=\sin^23x\)
1.Pt \(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=sin\left(x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\left(\dfrac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\\2x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Rightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\)\(\left(k\in Z\right)\)
2.\(sin^22x+cos^23x=1\)
\(\Leftrightarrow\dfrac{1-cos4x}{2}+\dfrac{1+cos6x}{2}=1\)
\(\Leftrightarrow cos6x=cos4x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow x=\dfrac{k\pi}{5}\)\(\left(k\in Z\right)\) (Gộp nghiệm)
Vậy...
3. \(Pt\Leftrightarrow\left(sinx+sin3x\right)+\left(sin2x+sin4x\right)=0\)
\(\Leftrightarrow2.sin2x.cosx+2.sin3x.cosx=0\)
\(\Leftrightarrow2cosx\left(sin2x+sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=-sin2x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\sin3x=sin\left(\pi+2x\right)\end{matrix}\right.\)(\(k\in Z\))
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\)(\(k\in Z\))\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\) (\(k\in Z\))
Vậy...
4. Pt\(\Leftrightarrow\dfrac{1-cos2x}{2}+\dfrac{1-cos4x}{2}=\dfrac{1-cos6x}{2}\)
\(\Leftrightarrow cos2x+cos4x=1+cos6x\)
\(\Leftrightarrow2cos3x.cosx=2cos^23x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\cosx=cos3x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=-k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy...
1) cos3x - cos4x + cos5x =0
2) sin3x + cos2x = 1 + 2sinx.cos2x
3) cos2x - cosx = 2 sin\(^2\)\(\dfrac{3x}{2}\)
4) cos\(^2\)2x + cos\(^2\)3x = sin\(^2\)x
5) sin3x.sin5x - cos4x.cos6x = 0
2.
\(sin3x+cos2x=1+2sinx.cos2x\)
\(\Leftrightarrow sin3x+cos2x=1+sin3x-sinx\)
\(\Leftrightarrow cos2x+sinx-1=0\)
\(\Leftrightarrow-2sin^2x+sinx=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
1.
\(cos3x-cos4x+cos5x=0\)
\(\Leftrightarrow cos3x+cos5x-cos4x=0\)
\(\Leftrightarrow2cos4x.cosx-cos4x=0\)
\(\Leftrightarrow\left(2cosx-1\right)cos4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{1}{2}\\cos4x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\4x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\end{matrix}\right.\)
3.
\(cos2x-cosx=2sin^2\dfrac{3x}{2}\)
\(\Leftrightarrow2sin\dfrac{3x}{2}.sin\dfrac{x}{2}+2sin^2\dfrac{3x}{2}=0\)
\(\Leftrightarrow2sin\dfrac{3x}{2}.\left(sin\dfrac{x}{2}+sin\dfrac{3x}{2}\right)=0\)
\(\Leftrightarrow sin\dfrac{3x}{2}.sinx.cos\dfrac{x}{2}=0\)
Đến đây dễ rồi tự làm tiếp nha.
Giải các phương trình sau:
1. tan2x+3= (1+√2 sin x)(tan x+ √2 cos x)
2. (1- cos x. cos2x )/ sin2x - 1/ cos x= 4 sin2x - sin x-1
3. sin3x + 2 cos3x+ cos2x - 2sin2x - 2sinx-1=0
a/\(\sin3x+\cos2x=1+2\sin x\cos2x\)
b/\(\sin^3x+\cos^3x=2\left(\sin^5x+\cos^5x\right)\)
c/\(\dfrac{\tan x}{\sin x}-\dfrac{\sin x}{\cos x}=\dfrac{\sqrt{2}}{2}\)
d/\(\dfrac{\cos x\left(\cos x+2\sin x\right)+3\sin x\left(\sin x+\sqrt{2}\right)}{\sin2x-1}=1\)
e/\(\sin^2x+\sin^23x-2\cos^22x=0\)
f/\(\dfrac{\tan x-\sin x}{\sin^3x}=\dfrac{1}{\cos x}\)
g/\(\sin2x\left(\cos x+\tan2x\right)=4\cos^2x\)
h/\(\sin^2x+\sin^23x=\cos^2x+\cos^23x\)
k/\(4\sin2x=\dfrac{\cos^2x-\sin^2x}{\cos^6x+\sin^6x}\)
mọi người giải giúp em với em đang cần gấp ạ
sin2x - cos5x = cos x- sin 6x
\(cos5x+cosx-\left(sin6x+sin2x\right)=0\)
\(\Leftrightarrow2cos3x.cos2x-2sin4x.cos2x=0\)
\(\Leftrightarrow2cos2x\left(cos3x-sin4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos3x=sin4x=cos\left(\dfrac{\pi}{2}-4x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\3x=\dfrac{\pi}{2}-4x+k2\pi\\3x=4x-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow...\)