# Solving Trigonometric Equations

1. Apr 5, 2007

### pugfug90

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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

2. Apr 5, 2007

### turdferguson

You can solve it like any other quadratic. What would you do with 2y2 + 3y = 0 ?

3. Apr 6, 2007

### mjsd

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. Apr 6, 2007

### Dick

As turdferguson points out, it's not completely valid to divide both sides by cos(x). What if cos(x)=0?

5. Apr 6, 2007

### pugfug90

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)?
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I think because adj/hyp, and adj has to be smaller than hyp, so cos or sin > 1 = no solution?

6. Apr 6, 2007

### Dick

Yes, that's the reason.

7. Apr 6, 2007

### HallsofIvy

Staff Emeritus
No real number solution. It is possible to have a complex value that has cosine equal to 1.5.

8. Apr 6, 2007

### Feldoh

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.