cos2x + cosx + 1 = 0
<=> 2cos^2 (x) - 1 - cosx + 1 = 0
<=> 2cos^2 (x) - cosx = 0
<=> cosx = 1/2 hoặc cosx = 0
<=> x = +- π/3 + k2π hoặc x = π/2 + kπ
cos2x + cosx + 1 = 0
<=> 2cos^2 (x) - 1 - cosx + 1 = 0
<=> 2cos^2 (x) - cosx = 0
<=> cosx = 1/2 hoặc cosx = 0
<=> x = +- π/3 + k2π hoặc x = π/2 + kπ
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
cosx + cos2x + cos3x + 1 = 0
sinx + sin2x + sin3x = 1 + cosx + cos2x
cos3x + sin3x + cosx - sinx = \(\sqrt{2}\)cos2x
sinx + sin2x + sin3x = cosx + cos2x + cos3x
cosx + cos2x + cos3x = 0
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
giải phương trình sau:
\(\dfrac{2sin^2x+cos4x-cos2x}{\left(sinx-cosx\right)sin2x}\)=0
Giải phương trình:
12sin3x+cos2x+cosx=0
Giải phương trình:
3sin2x + 2cos2x = 3
a) cos2x - sinx + cosx = 0
b) 2cos³x + sinx + cos2 = 0
sin2x-cosx=\(\sqrt{ }\)3(sinx+cos2x)