Question

How do you solve `2\sin ^ { 2} x \geq \sin x` if `0\leq x < 2\pi`?

Answer 1

intervals: `[pi/6, (5pi)/6] and [pi, 2pi]`

Explanation:

`f(x) = 2sin^2 x - sin x >= 0`
`f(x) = sin x(2sin x - 1) >= 0`
First solve f(x) = g(x).h(x) = 0
Solve the trig inequality algebraically by creating the sign chart of g(x) and h(x).
a. sin x = 0 --> x = 0 and `x = pi`, and `x = 2pi`
Create the variation chart of g(x) = sin x
g (x) is positive between `(0, pi)` and negative between `(pi, 2pi)`
b. `2sin x - 1 = 0` --> `sin x = 1/2`
Trig table and unit circle give:
`x = pi/6` and `x = (5pi)/6`
Create the variation chart of h(x) = 2sin x - 1
h(x) is positive inside (pi/6 and (5pi)/6).
f(x) has the resulting sign of g(x) and h(x).

x I 0 `pi/6` `(5pi)/6` `pi` `2pi`

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g(x) I 0 + + + 0 - 0

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h(x) I - 0 + 0 - -

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f(x) I - 0 + 0 - 0 + 0

From the chart, we see that f(x) > = 0 inside the intervals
[pi/6, (5pi)/6] and [pi, 2pi]

NOTE. The fastest way, if allowed, is to graph the trig function
f(x) = sin x(2sin x - 1). The parts of the graph that stay above the x-axis give the answers.