I have a question?

[sunrise]

Can you look where is my mistake?

2Cos(x) + 5Sec(x) =1.1 then cos(x)=?

2Cos(x)+5Sec(x) =11/10

2Cos(x)+5/Cos(x)=11/10

2Cos(x)*Cos(x)+5=11/10*Cos(x)

2Cos^2(x)+5=11/10*Cos(x)

2Cos^2(x)-11/10-Cos(x)-5=0 then i used delta=b^2-4ac

then i found delta=-4.79

X1=-b+sqrt(delta)/2a
then i found x1=2.19 and x2=-b-sqrt(delta)/2a but as i learnd answer is 0.5 where is the my mistake? Thank you

 

[girodd]

Can you look where is my mistake?

2Cos(x) + 5Sec(x) =1.1 then cos(x)=?

2Cos(x)+5Sec(x) =11/10

2Cos(x)+5/Cos(x)=11/10

2Cos(x)*Cos(x)+5=11/10*Cos(x)

2Cos^2(x)+5=11/10*Cos(x)

2Cos^2(x)-11/10-Cos(x)-5=0 then i used delta=b^2-4ac

then i found delta=-4.79

X1=-b+sqrt(delta)/2a
then i found x1=2.19 and x2=-b-sqrt(delta)/2a but as i learnd answer is 0.5 where is the my mistake? Thank you

A better way to go at this is to let cos(x) = u, sec(x) = 1/u, and solve for u.

 

[HallsofIvy]

But then you get 2u+ 5/u= 1.1 and multiplying both sides by u, \(\displaystyle 2u^2+ 5= 1.1u\) or \(\displaystyle 2u^2- 1.1u+ 5= 0\). The "discriminant" is (-1.1)^2- 4(2)(5)= 1.21- 40< 0 so there are no real solutions. Are you looking for complex number solutions?

 

[Deleted member 4993]

Looking at the answer, I think the problem should be:

2 * cos(x) + 5 * sec(x) = 11 ....................(instead of 1.1)