\(sin3x\left(cosx-sin3x\right)+cos3x\left(sinx-cos3x\right)=0\\ \Leftrightarrow sin3x\cdot cosx+cos3x\cdot sinx=sin^23x+cos^23x\\ \Leftrightarrow sin4x=1=sin\frac{\pi}{2}\\ \Leftrightarrow4x=\frac{\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{2}\)
sinx - sin3x + sin5x =0
sin2x + sin22x = sin23x
cos3x - cos5x = sinx
sin3x + sin5x + sin7x = 0
sinx + sin2x + sin3x - cosx - cos2x - cos3x = 0
chứng minh sin3x-cos3x=2(sinx+cosx)^3-3(sinx+cosx)
Giải các phương trình sau
a. Cosx+cos2x+cos3x+cos4x=0
b. Sinx+sin3x+sin5x+sin7x=0
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
sin3x - cos3x +sin2x+1=0
sin3x - cos3x +2sin2x +1 =0
•Sin3x - sin5x = sin2x
•Cosx + cos2x + cos3x = -1
•Sin2x + sin22x +sin23x + sin24x = 2
•1 + 2 sinxcos2x = sinx + cos2x
•Tan3x - tanx = sin2x
•(1-tanx)(1+sin2x) = 1+ tanx
\(\frac{ }{ }\)
tìm m sao cho (m-1)sin3x-(m+1)cos3x=2m-4
Giải các phương trình sau:
a, cos\(\left(3x-\frac{\pi}{6}\right)\)-sin \(\left(2x+\frac{\pi}{3}\right)\)=0
b, tan3x-tanx=0
c, cos2\(\left(x-\frac{\pi}{5}\right)\)=sin2\(\left(2x+\frac{4\pi}{5}\right)\)
d, 4cos2(2x-1)=0
e, cosx+cos2x+cos3x=0
f, 8sin2x.cos2x.cos4x=\(\sqrt{2}\)
g, cos3x-5cosx=sinx
h, sin7x-sin3x=cos5x
