How to solve the inequality 5sin x-2cos^2x-1>=0 in the interval [0,2pi]?

First, we'll express all inequality in terms of sin x. For
this reason, we'll apply the Pythagorean identity:

(cos
x)^2 = 1 - (sin x)^2

We'll re-write the
inequality:

5sin x - 2[1 - (sin x)^2] - 1
>=0

We'll remove the
brackets:

5sin x - 2 + 2(sin x)^2 - 1 >=
0

We'll combine like
terms:

2(sin x)^2 + 5sin x - 3 >=
0

We'll replace sin x by
t:

2t^2 + 5t - 3 >=
0

We'll determine the zeroes of the
quadratic:

t1 = [-5+sqrt(25 +
24)]/4

t1 = (-5 + 7)/4

t1 =
1/2

t2 = -3

sin x = t1
=> sin x = 1/2

The sine function is positive in the
1st and the 2nd quadrants and the values of x are:

x = pi/6
(1st quadrant)

x = pi - pi/6

x
= 5pi/6 (2nd quadrant)

sin x = t2 => sin x = -3
impossible since the value of sine function cannot be smaller than
-1.

We'll calculate the values of the inequality for x = 0,
x = pi/2 and x = 2pi.

f(0) = 2(sin 0)^2 + 5sin 0 -
3

f(0) = -3

f(pi/2) = 2(sin
pi/2)^2 + 5sin pi/2 - 3

f(pi/2) = 2 + 5 -
3

f(pi/2) = 4

f(2pi) = 2(sin
2pi)^2 + 5sin 2pi - 3

f(2pi) =
-3

We notice that the inequality is positive
over the interval [pi/6 ; 5pi/6].