The number of solution of the equation `sin^(3)x cos x+sin^(2)x cos^(2)x+cos^(3)x sin x=1` in the interval `[0, 2pi]` is equal to

Options

A. 0
B. 2
C. 4
D. 8

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 can follow these steps:

### Step 1: Rewrite the equation
The given equation can be rewritten as:
\[ \sin^3 x \cos x + \sin^2 x \cos^2 x + \cos^3 x \sin x = 1 \]

### Step 2: Factor out common terms
Notice that we can factor out \( \sin x \cos x \):
\[ \sin x \cos x (\sin^2 x + \cos^2 x) + \sin^2 x \cos^2 x = 1 \]
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we can simplify this to:
\[ \sin x \cos x + \sin^2 x \cos^2 x = 1 \]

### Step 3: Substitute \( \sin x \cos x \)
Let \( u = \sin x \cos x \). We know that \( \sin 2x = 2 \sin x \cos x \), thus:
\[ u = \frac{1}{2} \sin 2x \]
Substituting this into the equation gives:
\[ u + u^2 = 1 \]

### Step 4: Rearrange the equation
Rearranging the equation, we have:
\[ u^2 + u - 1 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 1, c = -1 \):
\[ u = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 6: Find the values of \( u \)
Thus, the solutions for \( u \) are:
\[ u_1 = \frac{-1 + \sqrt{5}}{2}, \quad u_2 = \frac{-1 - \sqrt{5}}{2} \]

### Step 7: Check the validity of \( u \)
Since \( u = \sin x \cos x \) must lie within the range \( [-\frac{1}{2}, \frac{1}{2}] \):
- For \( u_1 = \frac{-1 + \sqrt{5}}{2} \approx 0.618 \) (valid)
- For \( u_2 = \frac{-1 - \sqrt{5}}{2} \approx -1.618 \) (not valid)

### Step 8: Solve for \( x \)
Now we solve \( \sin x \cos x = \frac{-1 + \sqrt{5}}{2} \):
\[ \sin 2x = 2\sin x \cos x = 2 \cdot \frac{-1 + \sqrt{5}}{2} = -1 + \sqrt{5} \]

### Step 9: Determine the range of \( \sin 2x \)
Since \( \sqrt{5} \approx 2.236 \), we have:
\[ -1 + \sqrt{5} \approx 1.236 \]
This value is outside the range of the sine function, which is \([-1, 1]\).

### Conclusion
Since the only valid solution for \( u \) leads to an impossible value for \( \sin 2x \), we conclude that there are no solutions for the original equation in the interval \( [0, 2\pi] \).

Thus, the number of solutions is:
\[ \boxed{0} \]