Find the number of solutions of ` sin^(2) x cos^(2)x=1+cos^(2)x+sin^(4) x ` in the interval `[0,pi]`

Answer

To solve the equation \( \sin^2 x \cos^2 x = 1 + \cos^2 x + \sin^4 x \) in the interval \( [0, \pi] \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \sin^2 x \cos^2 x = 1 + \cos^2 x + \sin^4 x \]
We will rearrange this equation to bring all terms to one side:
\[ \sin^2 x \cos^2 x - \sin^4 x - \cos^2 x - 1 = 0 \]

### Step 2: Factor the equation
Next, we can factor out \( \cos^2 x \) from the left-hand side:
\[ \sin^2 x \cos^2 x - \sin^4 x - \cos^2 x - 1 = 0 \]
This can be rewritten as:
\[ 1 + \sin^4 x + \cos^2 x - \sin^2 x \cos^2 x = 0 \]
Now, we can group the terms:
\[ 1 + \sin^4 x + \cos^2 x - \sin^2 x \cos^2 x = 0 \]

### Step 3: Analyze the terms
Notice that:
- \( \sin^4 x \) is always non-negative.
- \( \cos^2 x \) is also always non-negative.
- \( \sin^2 x \cos^2 x \) is non-negative as well.

Thus, \( 1 + \sin^4 x + \cos^2 x \) is always greater than or equal to 1, since all terms are non-negative. Therefore, the left-hand side of the equation can never equal zero.

### Step 4: Conclusion
Since the left-hand side is always positive and can never be zero, we conclude that there are no solutions to the equation in the interval \( [0, \pi] \).

### Final Answer
The number of solutions of the equation \( \sin^2 x \cos^2 x = 1 + \cos^2 x + \sin^4 x \) in the interval \( [0, \pi] \) is **0**.

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