Solving Trigonometric Equations

  • Thread starter pugfug90
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  • #1
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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:
 

Answers and Replies

  • #2
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You can solve it like any other quadratic. What would you do with 2y2 + 3y = 0 ?
 
  • #3
mjsd
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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)
 
  • #4
Dick
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As turdferguson points out, it's not completely valid to divide both sides by cos(x). What if cos(x)=0?
 
  • #5
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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?
 
  • #6
Dick
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Yes, that's the reason.
 
  • #7
HallsofIvy
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No real number solution. It is possible to have a complex value that has cosine equal to 1.5.
 
  • #8
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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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