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

Options

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

Answer

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

### Step 1: Rearranging the Equation
We start with the given equation:
\[ \sin^3 x \cos x + \sin^2 x \cos^2 x + \cos^3 x \sin x + 1 = 0 \]
We can group the terms involving \(\sin x\) and \(\cos x\):
\[ \sin^3 x \cos x + \cos^3 x \sin x + \sin^2 x \cos^2 x + 1 = 0 \]

### Step 2: Factoring Common Terms
Notice that we can factor out \(\sin x \cos x\) from the first two terms:
\[ \sin x \cos x (\sin^2 x + \cos^2 x) + \sin^2 x \cos^2 x + 1 = 0 \]
Since \(\sin^2 x + \cos^2 x = 1\), we simplify this to:
\[ \sin x \cos x + \sin^2 x \cos^2 x + 1 = 0 \]

### Step 3: Substituting \(t = \sin x \cos x\)
Let \(t = \sin x \cos x\). Then, we have:
\[ t + t^2 + 1 = 0 \]
This can be rearranged as:
\[ t^2 + t + 1 = 0 \]

### Step 4: Solving the Quadratic Equation
Now we can solve the quadratic equation \(t^2 + t + 1 = 0\) using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = 1\), and \(c = 1\):
\[ t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} \]
Since the discriminant \(1 - 4 = -3\) is negative, the roots are imaginary.

### Step 5: Conclusion
Since \(t = \sin x \cos x\) has no real solutions (as it yields imaginary roots), there are no real values of \(x\) that satisfy the original equation in the interval \([0, 2\pi]\).

Thus, the number of solutions of the equation in the interval \([0, 2\pi]\) is:
\[ \boxed{0} \]