Answer :
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]
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]
