Solving an Equation in [-90,0]: Sec2A-3tanA-5=0

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Homework Help Overview

The problem involves solving the equation sec²A - 3tanA - 5 = 0 for A in the interval [-90, 0]. The context is trigonometric equations and their manipulation.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to manipulate the equation but expresses uncertainty about their approach. Some participants suggest rewriting sec²A in terms of tangent to facilitate solving the equation.

Discussion Status

The discussion includes attempts to clarify the manipulation of trigonometric identities. Guidance has been offered regarding the substitution of sec²A with an expression involving tangent, indicating a productive direction in the conversation.

Contextual Notes

Participants are working within the constraints of the specified interval for A and are exploring the implications of trigonometric identities in their attempts to solve the equation.

DERRAN
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Homework Statement


Solve the Equation:

sec2A - 3tanA - 5=0, for A [tex]\in[/tex][-90,0]

Homework Equations





The Attempt at a Solution


[tex]\frac{1}{cos^{2}A}[/tex] - [tex]\frac{3sinA}{cosA}[/tex] -5 =0

[tex]\frac{1-3sinA.cosA}{cos^{2}A}[/tex] -5 = 0

?
Did I go wrong? If not I'm stuck. Please help. Thank you.
 
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You didn't go wrong, but the form you wrote it in only makes it harder. I suggest you write the [itex]\sec^2 A[/itex] term in terms of the tangent. Do you know how?
 
Last edited:
[tex]sec^2A=\frac{1}{cos^2A}=\frac{sin^2A+cos^2A}{cos^2A}=\frac{sin^2A}{cos^2A}+\frac{cos^2A}{cos^2A}=tan^2A+1[/tex]

Just substitute and solve the quadratic equation.

Regards.
 
thanks
 

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