MHB Solving a Trigonometric Equation for Cos (x)

AI Thread Summary
The discussion centers on solving the trigonometric equation 2cos(x) + 5sec(x) = 1.1. A participant initially misapplies the equation, leading to confusion over the correct formulation and resulting in a negative discriminant. Clarifications reveal that the original equation was misquoted, affecting the calculations. The correct approach involves rewriting the equation in quadratic form, allowing for the application of the quadratic formula. The conversation emphasizes careful transcription of equations to avoid errors in solving trigonometric problems.
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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