Question

What is the solution to the trigonometric inequality 2sin(x)+3>sin^2(x) over the interval 0<=x<=2pi radians?

Answer 1

2 sin(x) + 3 > sin²(x)

can be rewritten as

sin²(x) - 2 sin(x) - 3 < 0

(sin(x) - 3) (sin(x) + 1) < 0

The left side is equal to zero when either

sin(x) - 3 = 0   or   sin(x) + 1 = 0

sin(x) = 3   or   sin(x) = -1

The first case can never happen for real x, since |sin(x)| ≤ 1. From the second case, we get one zero in the interval 0 ≤ x ≤ 2π at x = 3π/2.

Now, split up the solution domain into two intervals, and check the sign of the left side of the inequality at some point in the interval. If the inequality holds up, that interval will belong to the solution set for the inequality.

• From 0 ≤ x < 3π/2, take x = 0. Then

sin²(0) - 2 sin(0) - 3 = 0² - 0 - 3 < 0

is true.

• From 3π/2 < x ≤ 2π, take x = 2π. Then

sin²(2π) - 2 sin(2π) - 3 = 0² - 0 - 3 = -3 < 0

is also true.

This means both intervals 0 ≤ x < 3π/2 and 3π/2 < x ≤ 2π satisfy the inequality.