Solving a Trigonometric Equation for Cos (x)

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Discussion Overview

The discussion revolves around solving the trigonometric equation involving cosine and secant functions: "2cos(x) + 5sec(x) = 1.1". Participants are attempting to clarify the steps taken in solving the equation and identify any mistakes made in the process.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents their solution attempt but expresses confusion about arriving at an incorrect answer, suggesting a potential mistake in their calculations.
  • Another participant questions the accuracy of the problem statement, noting a discrepancy between the original equation and the subsequent steps taken, specifically regarding the transition from "cos(x)" to "2cos(x)".
  • A third participant confirms the original problem as "2cos(x) + 5sec(x) = 1.1" and seeks assistance in solving it.
  • A fourth participant points out the need to clarify whether the initial equation was misquoted or if it was a typographical error, suggesting a multiplication step to simplify the equation into a quadratic form.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the original problem statement, with some believing it was misquoted while others assert it is correct. The discussion remains unresolved regarding the correct approach to solving the equation.

Contextual Notes

There are unresolved issues regarding the accuracy of the problem statement and the calculations performed by the participants, particularly in transitioning between different forms of the equation.

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

Correct but can you help to solution?
 
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.

[math]2\cos(x)+5\sec(x)=1.1[/math]

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

[math]2\cos^2(x)+5=1.1\cos(x)[/math] or [math]2\cos^2(x)-1.1\cos(x)+5=0[/math]

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

[math]2u^2-1.1u+5=0[/math]

Can you make progress from here?
 

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