- #1
Karma
- 76
- 0
The question is solve [tex]2 + cos2x=3cosx[/tex] for 0<- X <-2pi greater/equal to zero and less than or equal to 2pi.
My solution
[tex]2+cos2x=3cosx[/tex]
[tex]2+cosx^2-sin^2=3cosx[/tex]
[tex]2+cosx^2-(1-cosx^2)=3cosx[/tex]
[tex]2+2cosx^2-1-3cosx=0[/tex]
[tex]2cos^2-3cosx+1=0[/tex]
Factoring
[tex]2x^2-3x+1=0[/tex]
[tex](2x-1)(x-1)=0[/tex]
[tex]x=1 or x=1/2[/tex]
solving for cos i get
[tex]cos(x)=1 --> x=0[/tex]
[tex]cos(x)=1/2 -->x= pi/3 or 5pi/3[/tex]
Did i do this Right?
My solution
[tex]2+cos2x=3cosx[/tex]
[tex]2+cosx^2-sin^2=3cosx[/tex]
[tex]2+cosx^2-(1-cosx^2)=3cosx[/tex]
[tex]2+2cosx^2-1-3cosx=0[/tex]
[tex]2cos^2-3cosx+1=0[/tex]
Factoring
[tex]2x^2-3x+1=0[/tex]
[tex](2x-1)(x-1)=0[/tex]
[tex]x=1 or x=1/2[/tex]
solving for cos i get
[tex]cos(x)=1 --> x=0[/tex]
[tex]cos(x)=1/2 -->x= pi/3 or 5pi/3[/tex]
Did i do this Right?