What is the solution to the trigonometric inequality 2sin (x) + 3> sin^2 (x)

First of all, let's analyze this equation. There is no restriction for

x since the domain of a function sin

(x)

is all real numbers.

Secondly, we can introduce an intermediary variable y = sin (x

and express this equation in terms of y : y 2 2 y + 3 or y 2 - 2 y 3 = 0

This equation has solutions: y 1, 2 = 2 ± √ 4 + 12 2 or y 1 = 3 and 2 = - 1

Now we have to take into consideration that y = sin (x)

and, as such, is restricted in values from - 1 to 1

(inclusive).

Therefore, solution y 1 = 3

would not lead to any value of x.

The second solution y 2 - 1

can be used to find x : sin (x) = 1

implies that x = - π 2 + 2 π N, where N - any integer number or, in degrees, x = -

90 o +

360 o ⋅

N

This is a final solution to the problem