The intervals that satisfy the given trigonometric Inequality are; 0 ≤ x < 3π/2 and 3π/2 < x ≤ 2π
<h3>How to solve trigonometric inequality?</h3>
We are given the trigonometric Inequality;
2 sin(x) + 3 > sin²(x)
Rearranging gives us;
sin²(x) - 2 sin(x) - 3 < 0
Factorizing this gives us;
(sin(x) - 3)(sin(x) + 1) < 0
Thus;
sin(x) - 3 = 0 or sin(x) + 1 = 0
sin(x) = 3 or sin(x) = -1
sin(x) = 3 is not possible because sin(x) ≤ 1.
Thus, we will work with;
sin(x) = -1 for the interval 0 ≤ x ≤ 2π radians.
Then, x = sin⁻¹(-1)
x = 3π/2.
Now, if we split up the solution domain into two intervals, we have;
from 0 ≤ x < 3π/2, at x = 0. Then;
sin²(0) - 2 sin(0) - 3
= 0² - 0 - 3
= -3 < 0
Thus, the interval 0 ≤ x < 3π/2 is true.
From 3π/2 < x ≤ 2π, take x = 2π. Then;
sin²(2π) - 2 sin(2π) - 3
= 0² - 0 - 3
= -3 < 0
Thus, the interval 3π/2 < x ≤ 2π is also true.
Read more about trigonometric inequality at; brainly.com/question/27862380
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