2(sin^2)(x) >= sin x

I need to solve the inequality 2*sin^2(x) >=(greater than or equal to) sin x on the interval 0<=x<2*pi. I have no idea where to start or how to solve this, could anybody help me out please and show me the steps?

2sin^2(x) ≥ sin(x)
Over interval 0 ≤ x < 2π

2sin^2(x)/sin(x) ≥ sin(x)/sin(x)

2sin(x) ≥ 1

sin(x) ≥ 1/2

Look at this table:
http://math2.org/math/graphs/unitcircle.…

π/6 ≤ x ≤ 5π/6

Short, clean, but incorrect... this I can assure you!

Click link below and check for the graph:
http://i.imgur.com/bQWNT.jpg

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Always remember rules in inequality, you mustn't cancel any term in the expression, just like how he did cancelling the sin{x} term.

Good luck!

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By the way, using multi account to thumb me down and to improve self rating WILL NOT change the fact that my solution is the only correct answer.

Looks like nobody has any CONSTRUCTIVE comment to add on what I've written. Oh well... .

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2sin²{x} ≥ sin{x}
2sin²{x} - sin{x} ≥ 0
(sin{x})(2sin{x} - 1) ≥ 0

The boundary values is when sin{x} = 0 and also 2sin{x} - 1 = 0.

When sin{x} = 0 ,
x = arcsin{0}
x = {0 , π} ...for the interval 0 ≤ x < 2π

When 2sin{x} - 1 = 0
sin{x} = 1/2
x = arcsin{1/2}
x = {π/6 , 5π/6} ...for the interval 0 ≤ x < 2π

Draw number line for boundaries at [0], [π/6], [5π/6], [π], [2π]; and then check for the inequality above.

.... [0] ...FALSE... [π/6] ...TRUE... [5π/6] ...FALSE... [π] ...TRUE... [2π] ....


For interval 0 ≤ x < 2π, the inequality given is only valid if
(π/6) ≤ x ≤ (5π/6) , or (π) ≤ x < (2π) .


Answer: (π/6) ≤ x ≤ (5π/6) ; or (π) ≤ x < (2π)
i.e. [π/6 , 5π/6] U [π , 2π)

1. divide by sin x on both sides: 2sin^2 x>= sin x ---->2sin x>=1
2. divide by 2: 2sin x>=1------->sin x>=.5
3. inverse sin: sin x>=.5-------->x>= sin^-1 .5
x>=30 degrees and 150 degrees
*it shows up as just 30 but it also works at 150 because sine means the height and on a unit circle, the height is the same on both halves of the y axis

2sin^2 x - sinx >= 0 ;

sinx(2sinx - 1) >= 0 ;

sinx>= 0 ; sinx >= 1/2.