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

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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].**