To solve the trigonometric inequality \(\sin^2(x) > 1 + \cos(x)\) over the interval \(0 \leq x \leq 2\pi\), let's follow a step-by-step approach.
1. Rewriting the Inequality:
Using the Pythagorean identity for sine squared, \(\sin^2(x) = 1 - \cos^2(x)\), we can rewrite the inequality:
[tex]\[
1 - \cos^2(x) > 1 + \cos(x)
\][/tex]
2. Simplifying the Inequality:
Next, we subtract 1 from both sides:
[tex]\[
1 - \cos^2(x) - 1 > \cos(x)
\][/tex]
Simplifying further, we get:
[tex]\[
-\cos^2(x) > \cos(x)
\][/tex]
Move all terms to one side to obtain:
[tex]\[
-\cos^2(x) - \cos(x) > 0
\][/tex]
3. Factoring the Expression:
We can factor out \(-\cos(x)\):
[tex]\[
\cos(x) (\cos(x) + 1) < 0
\][/tex]
4. Analyzing the Inequality:
The product \(\cos(x) (\cos(x) + 1)\) is less than 0, meaning one factor must be positive and the other negative. We analyze where \(\cos(x)\) and \(\cos(x) + 1\) have opposite signs:
- \(\cos(x)\) is negative when \( \frac{\pi}{2} < x < \frac{3\pi}{2} \).
- \(\cos(x) + 1\) is negative when \(\cos(x) < -1\), which is out of the range, so this doesn't affect our interval.
Therefore, \(\cos(x) (\cos(x) + 1)\) is negative when:
[tex]\[
\frac{\pi}{2} < x < \pi
\][/tex]
and
[tex]\[
\pi < x < \frac{3\pi}{2}
\][/tex]
5. Combining the Intervals:
Combining the valid intervals, we get:
[tex]\[
\frac{\pi}{2} < x < \pi \quad \text{and} \quad \pi < x < \frac{3\pi}{2}
\][/tex]
The solution to the inequality \(\sin^2(x) > 1 + \cos(x)\) over the interval \(0 \leq x \leq 2\pi\) is the combined intervals:
[tex]\[
\frac{\pi}{2} < x < \pi \quad \text{and} \quad \pi < x < \frac{3\pi}{2}
\][/tex]
So, the correct answer is:
[tex]\[
\boxed{\frac{\pi}{2} < x < \pi \text{ and } \pi < x < \frac{3\pi}{2}}
\][/tex]
