Trignometry Inequalities in [0, 2pi)?

[zeion]

Homework Statement

Solve the following equations or inequalities in the interval [0, 2pi).

2sin²x - 5sinx + 3 < 0

Homework Equations

The Attempt at a Solution

(2sin²x - 3)(sinx - 1)
sinx = 3/2 or sinx = 1

Not sure what to do now

sinx = 3/2 is impossible?
sinx = 1 then x = 90 = pi/2

 

[lanedance]

just correcting your square...

2sin²x - 5sinx + 3 = (2sinx-3)(sinx-1)

However rather than finding the bounding points (=0, though your thinking was correct) you have an inequality
(2sinx-3)(sinx-1) < 0

so what conditions need to be satisfied for (2sinx-3)(sinx-1) to be negative?

 

[zeion]

Do I need to sub in points to see rather it is positive or negative before and after sinx - 1?

 

[Mark44]

Your inequality in factored form is (2sinx-3)(sinx-1) < 0.
As you have already noticed, the first factor can't be zero, which means that it is either always positive or always negative, no matter what x value you substitute. Determine which of these it is.

For the product of the two factors to be negative, they have to be opposite in sign.

 

[lanedance]

following on from what Mark said, if the sin's are confusing, first solve for y, ie
(2y-3)(y-1)<0

then translate that to the original problem, by y = sinx, knowing that only y values in the range [-1,1] are allowable solutions for x

 

[zeion]

So I think that (2sinx-3) is always neagative, this means that I need to find values of x where (sinx -1) is positive to satisfy the inequality?

 

[lanedance]

sounds like you're on the right track to me

 

[zeion]

Does this mean I need to solve sinx > 1?
I don't think there are any values where sinx > 1. Or am I confused about what I'm doing?
sinx is 1 when x is 90

 

[Mark44]

sin(x) = 1 when x = (90 + k*360) degrees, but are there any values of x for which sin(x) > 1? If you're not sure, see lanedance's post 5.

 

[zeion]

What happens if there are no values where sinx > 1?

 

[Bohrok]

Then the inequality has no solutions.