\(2\cos^3x+\cos2x+\sin x=0\)
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
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.
Chứng minh
a) \(\dfrac{1+\cos x+\cos2x+\cos3x}{2\cos^2x+\cos x-1}=2\cos x\)
b) \(\cos\dfrac{5x}{2}.\cos\dfrac{3x}{2}+\sin\dfrac{7x}{2}.\sin\dfrac{x}{2}=\cos x.\cos2x\)
a, \(\dfrac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}\)
\(=\dfrac{1+cos2x+cosx+cos3x}{2cos^2x+cosx-1}\)
\(=\dfrac{2cos^2x+2cos2x.cosx}{cos2x+cosx}\)
\(=\dfrac{2cosx\left(cos2x+cosx\right)}{cos2x+cosx}=2cosx\)
b) \(cos\dfrac{5x}{2}.cos\dfrac{3x}{2}+sin\dfrac{7x}{2}.sin\dfrac{x}{2}\)
\(=cos\dfrac{4x+x}{2}.cos\dfrac{4x-x}{2}+sin\dfrac{4x+3x}{2}.sin\dfrac{4x-3x}{2}\)
\(=\dfrac{1}{2}\left(cos4x+cosx\right)-\dfrac{1}{2}\left(cos4x-cos3x\right)\)
\(=\dfrac{1}{2}\left(cosx+cos3x\right)=\dfrac{1}{2}.2cos2x.cos\left(-x\right)\)\(=cosx.cos2x\)
Cau1 :Nghiem cua phuong trinh cos2x+sin(x+pi/4)=0
Cau 2 ngiem cua phuong trinh sin(3x-5pi/6)+cos(3x+3pi/6)=0
1.
\(\cos2x+\sin\left(x+\frac{pi}{4}\right)=0\)
\(\Leftrightarrow\sin\left(x+\frac{pi}{4}\right)=-\cos2x\)
\(\Leftrightarrow\sin\left(x+\frac{pi}{4}\right)=\sin\left(2x-\frac{pi}{2}\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{pi}{4}=2x-\frac{pi}{2}+k2pi\\x+\frac{pi}{4}=pi-2x+\frac{pi}{2}+k2pi\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-x=-\frac{3}{4}pi+k2pi\\3x=+\frac{5}{4}pi+k2pi\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}pi+k2pi\\x=\frac{5}{12}pi+k\frac{2}{3}pi\end{cases}}\)
2.
\(\sin\left(3x-\frac{5pi}{6}\right)+\cos\left(3x+\frac{3pi}{6}\right)=0\)
\(\Leftrightarrow\sin\left(3x-\frac{5pi}{6}\right)=-\cos\left(3x+\frac{3pi}{6}\right)\)
\(\Leftrightarrow\sin\left(3x-\frac{5pi}{6}\right)=\sin\left(3x+\frac{3pi}{6}-\frac{pi}{2}\right)\)
\(\Leftrightarrow\orbr{\begin{cases}3x-\frac{5pi}{6}=3x+\frac{3pi}{6}-\frac{pi}{2}+k2pi\\3x-\frac{5pi}{6}=pi-3x-\frac{3pi}{6}+\frac{pi}{2}+k2pi\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}0x=\frac{5pi}{6}+k2pi\left(VN\right)\\6x=\frac{11pi}{6}+k2pi\end{cases}}\)
\(\Leftrightarrow x=\frac{11pi}{36}+k\frac{1}{3}pi\)
giải pt:
\(\frac{sin^3x+\cos^3x}{2\cos x-\sin x}=\cos2x\)
Giải các phương trình sau:
1, \(\sin2x.\sin x+\cos2x=\cos2x.\cos x\)
2, \(\frac{\tan x+1}{1-\tan x}=\cot2x\)
3, \(\sin3x+\sin x-\sqrt{3}\cos x=0\)
4, \(\sin^3x-\cos^3x-\sin x.\cos x=1\)
5, \(8\sin x-\frac{1}{\sin x}=\frac{\sqrt{3}}{\cos x}\)
Mọi người giúp em với ạ!!! Em cần gấp!!!
Giải các phương trình sau :
a) \(\sin\left(x-\frac{2\pi}{3}\right)=\cos2x\)
b) \(\frac{\sin^3x+\cos^3x}{2\cos x-\sin x}=\cos2x\)
a) Ta có : \(sin\left(x-\frac{2\pi}{3}\right)=cos2x\)
\(\Leftrightarrow sin\left(x-\frac{2\pi}{3}\right)=sin\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{2\pi}{3}=\frac{\pi}{2}-2x+k2\pi\\x-\frac{2\pi}{3}=\pi-\frac{\pi}{2}+2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{18}+k\frac{2\pi}{3}\\x=-\frac{7\pi}{6}-k2\pi\end{matrix}\right.\)
Vậy ...
giải phương trình \(\frac{\sin^3x-\cos^3x}{\sqrt{\sin x}+\sqrt{\cos x}}=2\cos2x\)
Lời giải:
ĐKXĐ: ...............
PT \(\Leftrightarrow \frac{(\sin x-\cos x)(\sin ^2x+\sin x\cos x+\cos ^2x)}{\sqrt{\sin x}+\sqrt{\cos x}}=-2(\sin x-\cos x)(\sin x+\cos x)\)
\(\Leftrightarrow (\sin x-\cos x)\left[\frac{\sin ^2x+\sin x\cos x+\cos ^2x}{\sqrt{\sin x}+\sqrt{\cos x}}+2(\sin x+\cos x)\right]=0\)
Dễ thấy với $\sin x, \cos x\geq 0$ thì biểu thức trong ngoặc vuông luôn lớn hơn $0$
Do đó:
$\sin x-\cos x=0$
$\Leftrightarrow \sin x=\cos x$
Mà $\sin ^2x+\cos ^2x=1; \sin x, \cos x\geq 0$ nên $\sin x=\cos x=\frac{1}{\sqrt{2}}$
$\Rightarrow x=k\pi -\frac{7}{4}\pi$ với $k$ nguyên.
giải các phương trình sau :
1. sin( x+\(\pi\)/4)=2/3
2.cos2x-5sinx-3=0
3.cos3x=sin2x
4.cos3x=-\(\sqrt{ }\)3 với -\(\pi\)/2<x<0
5.4sin\(^4\)x + 12cos\(^2\)x=7
6.cot(x-1)=(cos2x)/(1+tanx) + sin\(^2\)x - 1/2sin2x
7.sin\(^2\)3x-cos\(^2\)4x=sin\(^2\)5x-cos\(^2\)6x