hey does anyone know how to solve from 0≦@≦2pi 4cos2x=3sec2x?
jemsta I sit here alone Joined Apr 6, 2005 Messages 5,711 Location O.P Gender Male HSC 2005 Oct 25, 2005 #1 hey does anyone know how to solve from 0≦@≦2pi 4cos2x=3sec2x?
W word. Member Joined Sep 20, 2005 Messages 174 Gender Male HSC 2005 Oct 25, 2005 #2 multiply both sides by cos2x divide both sides by 4 square root both sides
B Bellow Member Joined Oct 3, 2005 Messages 36 Gender Male HSC 2005 Oct 25, 2005 #3 azn_gangsta81 said: hey does anyone know how to solve from 0≦@≦2pi 4cos2x=3sec2x? Click to expand... 4cos2x=3sec2x 4cos2x=3/cos2x 4cos²2x=3 cos²2x=3/4 cos2x=+-√3/2 2x=π/6, 5π/6, 7π/6, 11π/6, 13π/6, 17π/6, 19π/6 23π/6 x=π/12, 5π/12, 7π//12, 11π/12, 13π/12, 17π/12, 19π/12, 23π/12 Last edited: Oct 25, 2005
azn_gangsta81 said: hey does anyone know how to solve from 0≦@≦2pi 4cos2x=3sec2x? Click to expand... 4cos2x=3sec2x 4cos2x=3/cos2x 4cos²2x=3 cos²2x=3/4 cos2x=+-√3/2 2x=π/6, 5π/6, 7π/6, 11π/6, 13π/6, 17π/6, 19π/6 23π/6 x=π/12, 5π/12, 7π//12, 11π/12, 13π/12, 17π/12, 19π/12, 23π/12
jemsta I sit here alone Joined Apr 6, 2005 Messages 5,711 Location O.P Gender Male HSC 2005 Oct 25, 2005 #4 thanks buddy
D damo676767 Member Joined Oct 20, 2004 Messages 149 Location Winmalee, Blue Mountains Gender Male HSC 2005 Oct 25, 2005 #5 Bellow said: 4cos2x=3sec2x 4cos2x=3/cos2x 4cos²2x=3 cos²2x=3/4 cos2x=√3/2 2x=π/6, 11π/6, 13π/6, 23π/6 x=π/12, 11π/12, 13π/12, 23π/12 Click to expand... you left of values by not concidering cos2x=- √3/2 eg 5π/12
Bellow said: 4cos2x=3sec2x 4cos2x=3/cos2x 4cos²2x=3 cos²2x=3/4 cos2x=√3/2 2x=π/6, 11π/6, 13π/6, 23π/6 x=π/12, 11π/12, 13π/12, 23π/12 Click to expand... you left of values by not concidering cos2x=- √3/2 eg 5π/12
B Bellow Member Joined Oct 3, 2005 Messages 36 Gender Male HSC 2005 Oct 25, 2005 #6 damo676767 said: you left of values by not concidering cos2x=- √3/2 eg 5π/12 Click to expand... oh yer ill edit now
damo676767 said: you left of values by not concidering cos2x=- √3/2 eg 5π/12 Click to expand... oh yer ill edit now
