Bài 3: Một số phương trình lượng giác thường gặp
Các câu hỏi tương tự
sinx + sin2x + sin3x = 1 + cosx + cos2x
cos3x + sin3x + cosx - sinx = \(\sqrt{2}\)cos2x
sinx + sin2x + sin3x = cosx + cos2x + cos3x
Giúp mình với mn...
1)cos2x+cos22x+cos23x+cos24x=2
2) (1-tanx) (1+sin2x)=1+tanx
3) tan2x=sin3x.cosx
4) tanx +cot2x=2cot4x
5) sinx+sin2x+sin3x=cosx+cos2x+cos3x
6)sinx=√2 sin5x-cosx
7) 1/sin2x + 1/cos2x =2/sin4x
8) sinx+cosx=cos2x/1-sin2x
9)1+cos2x/cosx= sin2x/1-cos2x
10)sin3x+cos3x/2cosx-sinx=cos2x
26 tháng 7 2020 lúc 21:02
Giải các pt:
a) \(cos3x-sinx=\sqrt{3}\left(cosx-sin3x\right)\)
b) \(2cos^2x-3\sqrt{3}sin2x-4sin^2x=-4\)
c) \(\sqrt{3}\left(cos2x+sin3x\right)=sin2x+cos8x\)
d) \(cos2x-\sqrt{3}sin2x=\sqrt{3}sinx+cosx\)
e) \(sin8x-cos6x=\sqrt{3}\left(sin6x+cos8x\right)\)
cosx-2cos3x1+sqrt{3}sinx
sinx+sinxleft(x+dfrac{pi}{3}right)+sin4xsinleft(2x-dfrac{pi}{3}right)
left(1-dfrac{1}{2sinx}right)cos^22x2sinx-3+dfrac{1}{sinx}
( sinx -2cosx)cos2x + sinx (cos4x - 1)cosx +dfrac{cos2x}{2sinx}
left(dfrac{cos4x+sin2x}{cos3x+sin3x}right)^22sqrt{2}sinleft(x+dfrac{pi}{4}right)+3
Đọc tiếp
\(cosx-2cos3x=1+\sqrt{3}sinx\)
\(sinx+sinx\left(x+\dfrac{\pi}{3}\right)+sin4x=sin\left(2x-\dfrac{\pi}{3}\right)\)
\(\left(1-\dfrac{1}{2sinx}\right)cos^22x=2sinx-3+\dfrac{1}{sinx}\)
( sinx -2cosx)cos2x + sinx = (cos4x - 1)cosx +\(\dfrac{cos2x}{2sinx}\)
\(\left(\dfrac{cos4x+sin2x}{cos3x+sin3x}\right)^2=2\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+3\)
(cosx+sin2x)/(cos2x+sinx)=căn3
\(\dfrac{\sqrt{2}\left(sinx-cox\right)^2\left(1+2sin2x\right)}{sin3x+sin5x}=1-tanx\)
\(sin\left(2x-\dfrac{\pi}{4}\right)cos2x-2\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=0\)
(sin2x+cos2x)cosx+2cos2x -sinx=0
sinx + cosxsin2x + \(\sqrt{3}cos3x=2\left(cos4x+sin^3x\right)\)
\(\sqrt{3}cos5x-2sin3xcos2x-sinx=0\)
giải phương trình sau:
\(\dfrac{2sin^2x+cos4x-cos2x}{\left(sinx-cosx\right)sin2x}\)=0
1) giải pt:
a) cosx.cosx=cos2x.cos4x
b) cos5x.sin4x=cos3x.sin2x
c) sinx+sin2x=cosx+cos2x
d) sin2x+sin4x=sin6x
a/ 1-sin9x+\(\sqrt{3}\)cos9x=0
b/ \(\sqrt{3}\)sin2x-2sin2x=\(\sqrt{2}\)-1
c/ sin5x+\(\sqrt{3}\)cos5x=2sin7x
d/ cos2x+sinx=\(\sqrt{3}\)(cosx-sin2x)
e/ sin2x+4sin2x+3cos2x+2=0
f/ 2sin2x+cos2x=7sinx-2cosx+4