Solve the inequality ` (5)/4 sin^(2) x + sin^(2) x * cos^(2)x gt cos 2x. `

Answer

To solve the inequality \( \frac{5}{4} \sin^2 x + \sin^2 x \cos^2 x > \cos 2x \), we will follow these steps:

### Step 1: Rewrite \( \cos 2x \)
We know that:
\[ \cos 2x = 2 \cos^2 x - 1 \]
Thus, we can rewrite the inequality as:
\[ \frac{5}{4} \sin^2 x + \sin^2 x \cos^2 x > 2 \cos^2 x - 1 \]

### Step 2: Substitute \( \sin^2 x \)
Using the identity \( \sin^2 x = 1 - \cos^2 x \), we can substitute \( \sin^2 x \):
\[ \frac{5}{4} (1 - \cos^2 x) + (1 - \cos^2 x) \cos^2 x > 2 \cos^2 x - 1 \]

### Step 3: Simplify the left-hand side
Expanding the left-hand side:
\[ \frac{5}{4} - \frac{5}{4} \cos^2 x + \cos^2 x - \cos^4 x > 2 \cos^2 x - 1 \]
Combine like terms:
\[ \frac{5}{4} - \frac{5}{4} \cos^2 x + \cos^2 x - 2 \cos^2 x + 1 > -\cos^4 x \]
This simplifies to:
\[ \frac{9}{4} - \frac{9}{4} \cos^2 x > -\cos^4 x \]

### Step 4: Rearrange the inequality
Rearranging gives:
\[ \cos^4 x - \frac{9}{4} \cos^2 x + \frac{9}{4} > 0 \]

### Step 5: Let \( t = \cos^2 x \)
Let \( t = \cos^2 x \), then we have:
\[ t^2 - \frac{9}{4} t + \frac{9}{4} > 0 \]

### Step 6: Solve the quadratic inequality
To find the roots of the quadratic equation:
\[ t^2 - \frac{9}{4} t + \frac{9}{4} = 0 \]
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{\frac{9}{4} \pm \sqrt{\left(\frac{9}{4}\right)^2 - 4 \cdot 1 \cdot \frac{9}{4}}}{2 \cdot 1} \]
Calculating the discriminant:
\[ \left(\frac{9}{4}\right)^2 - 4 \cdot 1 \cdot \frac{9}{4} = \frac{81}{16} - \frac{36}{4} = \frac{81}{16} - \frac{144}{16} = -\frac{63}{16} \]
Since the discriminant is negative, the quadratic does not cross the x-axis and is always positive.

### Step 7: Determine the range of \( t \)
Since \( t = \cos^2 x \) must be between 0 and 1, we conclude:
\[ \cos^2 x \in (0, 1) \]

### Step 8: Find the values of \( x \)
From \( \cos^2 x = t \):
- \( \cos x = \sqrt{t} \) or \( \cos x = -\sqrt{t} \)
- This gives us:
\[ x = n\pi + \cos^{-1}(\sqrt{t}) \quad \text{and} \quad x = n\pi - \cos^{-1}(\sqrt{t}) \]

### Final Answer
The solution to the inequality is:
\[ x \in \left( n\pi + \frac{\pi}{6}, n\pi + \frac{5\pi}{6} \right) \quad \text{for integers } n \]