The solutions to the trigonometric inequality in the interval is given by:
[tex]\frac{\pi}{6} \leq x \leq \frac{5\pi}{6}[/tex], or [tex]\frac{7\pi}{6} \leq x \leq \frac{11\pi}{6}[/tex]
It is given by:
[tex]\sin^{2}{x} > \cos{x}[/tex]
We have the following identity:
[tex]\sin^2{x} = 2\sin{x}\cos{x}[/tex]
Hence:
[tex]2\sin{x}\cos{x} > \cos{x}[/tex]
[tex]2\sin{x} > 1[/tex]
[tex]\sin{x} > \frac{1}{2}[/tex]
Between the first and second quadrant, the solution is:
[tex]\frac{\pi}{6} \leq x \leq \frac{5\pi}{6}[/tex]
Between the third and fourth quadrants, the solution is:
[tex]\frac{7\pi}{6} \leq x \leq \frac{11\pi}{6}[/tex]
More can be learned about trigonometric equations at https://brainly.com/question/24680641
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