The number of solutions of the equation `sin^3x cos x + sin^2 x cos^2x +cos^3 x sin x = 1` in the interval `[0,2pi]` is /are

Options

A. 1
B. 2
C. 3
D. 0

Answer

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

### Step 1: Factor the Left Side
We can factor out \( \sin x \cos x \) from the left-hand side of the equation:
\[ \sin x \cos x (\sin^2 x + \sin x \cos x + \cos^2 x) = 1 \]

### Step 2: Simplify the Expression
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we can rewrite the expression inside the parentheses:
\[ \sin^2 x + \cos^2 x = 1 \implies \sin^2 x + \sin x \cos x + \cos^2 x = 1 + \sin x \cos x \]
Thus, the equation becomes:
\[ \sin x \cos x (1 + \sin x \cos x) = 1 \]

### Step 3: Substitute \( \sin x \cos x \)
Let \( u = \sin x \cos x \). The equation now reads:
\[ u(1 + u) = 1 \]
Expanding this gives:
\[ u + u^2 = 1 \implies u^2 + u - 1 = 0 \]

### Step 4: Solve the Quadratic Equation
We can solve this quadratic equation using the quadratic formula:
\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 5: Evaluate the Roots
Calculating the roots:
1. \( u_1 = \frac{-1 + \sqrt{5}}{2} \)
2. \( u_2 = \frac{-1 - \sqrt{5}}{2} \)

Calculating \( \sqrt{5} \approx 2.236 \):
- \( u_1 \approx \frac{-1 + 2.236}{2} \approx \frac{1.236}{2} \approx 0.618 \)
- \( u_2 \approx \frac{-1 - 2.236}{2} \approx \frac{-3.236}{2} \approx -1.618 \)

### Step 6: Check Validity of Roots
Since \( u = \sin x \cos x \) must lie in the range \( [-1, 1] \):
- \( u_1 \approx 0.618 \) is valid.
- \( u_2 \approx -1.618 \) is invalid (since it is less than -1).

### Step 7: Find \( x \) from \( u_1 \)
Now we need to find \( x \) such that:
\[ \sin x \cos x = \frac{-1 + \sqrt{5}}{2} \]
Using the identity \( \sin 2x = 2 \sin x \cos x \):
\[ \sin 2x = 2u = 2 \cdot \frac{-1 + \sqrt{5}}{2} = -1 + \sqrt{5} \]

### Step 8: Determine the Number of Solutions
The value \( -1 + \sqrt{5} \approx 1.236 \) is outside the range of the sine function, which is \( [-1, 1] \). Therefore, there are no solutions for \( \sin 2x = -1 + \sqrt{5} \).

### Conclusion
Thus, the number of solutions of the equation \( \sin^3 x \cos x + \sin^2 x \cos^2 x + \cos^3 x \sin x = 1 \) in the interval \( [0, 2\pi] \) is:
\[ \boxed{0} \]