Solving a Trigonometric Equation for Cos (x)

[melissax]

Hi, may you help me?
I have a question I solved but couldn't ended.

If cos(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

 

[Ackbach]

Hi, may you help me?
I have a question I solved but couldn't ended.

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

In-between these two lines, you have copied incorrectly. Where is the 2 coming from? Is that in the original problem or not?

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

You have made another error here: $(11/10)\cos(x)$ became $(11/10)-\cos(x)$.

You need to be more careful when you go from one line to the next. Pretend someone's holding a gun to your head and will fire if you make a mistake!

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

For the original problem, I get complex values for $\cos(x)$. Can you please post the original problem verbatim?

 

[melissax]

question is "2cos(x)+ 5sec(x)=1.1" problem isn't wrong.

Correct but can you help to solution?

 

[Jameson]

In your original post you wrote the problem as "cos(x)+5Sec(x)=1.1" then you switched the "cos(x)" to "2cos(x)" on the next line. We can't tell if this is a math mistake which we need to explain or just a typo. I'm going to guess it's a typo.

\(\displaystyle 2\cos(x)+5\sec(x)=1.1\)

Multiply everything by $\cos(x)$ and you get

\(\displaystyle 2\cos^2(x)+5=1.1\cos(x)\) or \(\displaystyle 2\cos^2(x)-1.1\cos(x)+5=0\)

This has a quadratic form to it. Let $u=\cos(x)$ and rewrite the above equation as

\(\displaystyle 2u^2-1.1u+5=0\)

Can you make progress from here?

 

[CellCoree]

Hello! It looks like you have made a small mistake in your calculations. When you used the quadratic formula to solve for x, you should have used 2a instead of a in the denominator. This would give you x1=1.28 and x2=-3.91. However, neither of these values satisfy the original equation of cos(x)+5sec(x)=1.1. This means that there is no real solution to this equation. In this case, the answer would be "no solution". I hope this helps!