trig equations (1 Viewer)

[jemsta]

hey does anyone know how to solve from 0≦@≦2pi
4cos2x=3sec2x?

 

[word.]

multiply both sides by cos2x
divide both sides by 4
square root both sides

 

[Bellow]

hey does anyone know how to solve from 0≦@≦2pi
4cos2x=3sec2x?

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]

thanks buddy

 

[damo676767]

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

you left of values by not concidering cos2x=- √3/2

eg 5π/12

 

[Bellow]

you left of values by not concidering cos2x=- √3/2

eg 5π/12

oh yer ill edit now