Find the number of solution of the equations
`sin^3 x cos x + sin^(2) x* cos^(2) x+sinx * cos^(3) x=1`, when ` x in[0,2pi]`

Answer

To solve the equation \( \sin^3 x \cos x + \sin^2 x \cos^2 x + \sin x \cos^3 x = 1 \) for \( x \) in the interval \( [0, 2\pi] \), we will follow these steps:

### Step 1: Factor out common terms
We can factor out \( \sin x \cos x \) from the left-hand side of the equation:

\[ \sin^3 x \cos x + \sin^2 x \cos^2 x + \sin x \cos^3 x = \sin x \cos x \left( \sin^2 x + \sin x \cos x + \cos^2 x \right) \]

### Step 2: Simplify the expression inside the parentheses
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we can rewrite the expression:

\[ \sin^2 x + \sin x \cos x + \cos^2 x = 1 + \sin x \cos x \]

Thus, our equation becomes:

\[ \sin x \cos x (1 + \sin x \cos x) = 1 \]

### Step 3: Multiply both sides by 4
To facilitate the use of double angle identities, we multiply both sides by 4:

\[ 4 \sin x \cos x (1 + \sin x \cos x) = 4 \]

### Step 4: Use the double angle identity
Recall that \( 2 \sin x \cos x = \sin 2x \). Therefore, we can express the left-hand side as:

\[ 2 \sin 2x (1 + \frac{1}{2} \sin 2x) = 4 \]

### Step 5: Rearranging the equation
This leads to:

\[ 2 \sin 2x + \sin^2 2x - 4 = 0 \]

### Step 6: Substitute \( u = \sin 2x \)
Let \( u = \sin 2x \). The equation becomes:

\[ u^2 + 2u - 4 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ u = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 16}}{2} = \frac{-2 \pm \sqrt{20}}{2} = \frac{-2 \pm 2\sqrt{5}}{2} = -1 \pm \sqrt{5} \]

### Step 8: Evaluate the roots
The roots are:

\[ u_1 = -1 + \sqrt{5} \quad \text{and} \quad u_2 = -1 - \sqrt{5} \]

### Step 9: Check the range of \( u \)
Since \( \sin 2x \) must lie in the range \([-1, 1]\):

1. For \( u_1 = -1 + \sqrt{5} \approx 1.236 \) (not in the range).
2. For \( u_2 = -1 - \sqrt{5} \approx -3.236 \) (not in the range).

### Conclusion
Both roots are outside the range of the sine function. Therefore, there are no solutions to the equation \( \sin^3 x \cos x + \sin^2 x \cos^2 x + \sin x \cos^3 x = 1 \) in the interval \( [0, 2\pi] \).