Solving Trigonometric Equations

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Homework Help Overview

The discussion revolves around solving the trigonometric equation \(2\cos^2 x + 3\cos x = 0\) within the interval \([0, 2\pi)\). Participants are exploring the use of inverse functions and graphing utilities to verify solutions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss factoring the equation and the implications of dividing by \(\cos x\). There is a focus on identifying valid solutions and the reasoning behind why \(\cos x = -\frac{3}{2}\) is not acceptable in the context of trigonometric functions.

Discussion Status

There is an ongoing exploration of the problem, with some participants suggesting methods for solving the equation and others questioning the validity of certain steps. The discussion includes considerations of potential solutions and the nature of the cosine function.

Contextual Notes

Participants note that the problem is part of a midterm and reference previous learning, indicating a potential gap in confidence regarding the material. There is also mention of the limitations of cosine values in relation to their definitions.

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Homework Statement


use the inverse functions where necessary to find all solution of the equation in the interval [0,2pi). Use a graphing utility to verify.

2cos(squared)x+3cosx=0


Homework Equations





The Attempt at a Solution



The answer is pi/2 and 3pi/2 from my teacher..
===
2cos(squared)x=-3cosx
2cosx=-3
cosx=-3/2

stuck :confused:
 
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You can solve it like any other quadratic. What would you do with 2y2 + 3y = 0 ?
 
also remember you cannot divide both side by cos x (or y in turdferguson redefinition), because cos x can be zero (and in fact it is a solution)
 
As turdferguson points out, it's not completely valid to divide both sides by cos(x). What if cos(x)=0?
 
I can factor out cosx? (sorry, this is midterm, and this stuff is what we did in January:p)

2cos(squared)x+3cosx=0
cosx(2cosx+3)=0
cosx=0 and cosx=-3/2

I can figure cosx=0 to be x=pi/2 and 3pi/2 but can anyone remind me why cosx=-3/2 doesn't work (invalid in calc)?
---
I think because adj/hyp, and adj has to be smaller than hyp, so cos or sin > 1 = no solution?
 
Yes, that's the reason.
 
No real number solution. It is possible to have a complex value that has cosine equal to 1.5.
 
Could you also look at it this way? f(x)=2cox(x)+3 -- It's going to be translated? up 3 units and is only has an amplitude of 2 so it will never cross the x-axis, so there are not any x-intercepts.
 

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