Solve the inequality `sin2x gtsqrt(2)sin^2x+(2-sqrt(2))cos^2x`

Answer

To solve the inequality \( \sin 2x > \sqrt{2} \sin^2 x + (2 - \sqrt{2}) \cos^2 x \), we can follow these steps:

### Step 1: Rewrite the inequality
We start with the given inequality:
\[ \sin 2x > \sqrt{2} \sin^2 x + (2 - \sqrt{2}) \cos^2 x \]
Recall that \( \sin 2x = 2 \sin x \cos x \). Therefore, we can rewrite the inequality as:
\[ 2 \sin x \cos x > \sqrt{2} \sin^2 x + (2 - \sqrt{2}) \cos^2 x \]

### Step 2: Substitute \( \cos^2 x \)
Using the identity \( \cos^2 x = 1 - \sin^2 x \), we can substitute \( \cos^2 x \) in the inequality:
\[ 2 \sin x \cos x > \sqrt{2} \sin^2 x + (2 - \sqrt{2})(1 - \sin^2 x) \]
Expanding the right-hand side:
\[ 2 \sin x \cos x > \sqrt{2} \sin^2 x + (2 - \sqrt{2}) - (2 - \sqrt{2}) \sin^2 x \]
Combining like terms:
\[ 2 \sin x \cos x > (2 - 2\sin^2 x) + (2 - \sqrt{2}) - \sqrt{2} \sin^2 x \]
This simplifies to:
\[ 2 \sin x \cos x > 2 - (2 + \sqrt{2}) \sin^2 x \]

### Step 3: Rearranging the inequality
Rearranging gives us:
\[ 2 \sin x \cos x + (2 + \sqrt{2}) \sin^2 x - 2 > 0 \]

### Step 4: Factor the left-hand side
Let \( y = \sin x \). Then we can rewrite the inequality:
\[ 2y\sqrt{1 - y^2} + (2 + \sqrt{2})y^2 - 2 > 0 \]
This is a quadratic inequality in terms of \( y \).

### Step 5: Analyze the quadratic
To analyze the quadratic inequality, we can find the roots and determine the intervals where the inequality holds.

### Step 6: Solve for critical points
Set the left-hand side equal to zero to find critical points:
\[ 2y\sqrt{1 - y^2} + (2 + \sqrt{2})y^2 - 2 = 0 \]
This may require numerical methods or graphing to find the exact points.

### Step 7: Test intervals
Once you have the critical points, test intervals around these points to determine where the inequality holds true.

### Step 8: Conclusion
The solution will be the intervals where the inequality is satisfied.