The number of solution of the equation `2 "sin"^(3) x + 2 "cos"^(3) x - 3 "sin" 2x + 2 = 0 "in" [0, 4pi]`, is

Options

A. 2
B. 3
C. 4
D. 5

Explanation

To solve the equation \( 2 \sin^3 x + 2 \cos^3 x - 3 \sin 2x + 2 = 0 \) in the interval \( [0, 4\pi] \), we will follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that \( \sin 2x = 2 \sin x \cos x \). Substituting this into the equation gives:
\[ 2 \sin^3 x + 2 \cos^3 x - 3(2 \sin x \cos x) + 2 = 0 \]
This simplifies to:
\[ 2 \sin^3 x + 2 \cos^3 x - 6 \sin x \cos x + 2 = 0 \]

### Step 2: Factor the equation
We can factor \( 2 \sin^3 x + 2 \cos^3 x \) using the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \):
\[ 2(\sin^3 x + \cos^3 x) = 2(\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x) \]
Since \( \sin^2 x + \cos^2 x = 1 \), we have:
\[ \sin^2 x - \sin x \cos x + \cos^2 x = 1 - \sin x \cos x \]
Thus, the equation becomes:
\[ 2(\sin x + \cos x)(1 - \sin x \cos x) - 6 \sin x \cos x + 2 = 0 \]

### Step 3: Simplify the equation
Now, we can rearrange the equation:
\[ 2(\sin x + \cos x)(1 - \sin x \cos x) + 2 - 6 \sin x \cos x = 0 \]
This can be simplified further, but let's analyze the roots directly.

### Step 4: Set up the conditions for solutions
We can set:
\[ \sin x + \cos x = 0 \quad \text{or} \quad 1 - \sin x \cos x = 0 \]

### Step 5: Solve for \( \sin x + \cos x = 0 \)
The equation \( \sin x + \cos x = 0 \) implies:
\[ \tan x = -1 \]
The solutions for \( x \) in the interval \( [0, 4\pi] \) are:
\[ x = \frac{3\pi}{4}, \frac{7\pi}{4}, \frac{11\pi}{4}, \frac{15\pi}{4} \]

### Step 6: Solve for \( 1 - \sin x \cos x = 0 \)
This implies:
\[ \sin x \cos x = 1 \quad \Rightarrow \quad \frac{1}{2} \sin 2x = 1 \quad \Rightarrow \quad \sin 2x = 2 \]
This has no solutions since the sine function cannot exceed 1.

### Step 7: Count the total number of solutions
The only valid solutions come from \( \sin x + \cos x = 0 \). We found four solutions:
1. \( x = \frac{3\pi}{4} \)
2. \( x = \frac{7\pi}{4} \)
3. \( x = \frac{11\pi}{4} \)
4. \( x = \frac{15\pi}{4} \)

Thus, the total number of solutions in the interval \( [0, 4\pi] \) is **4**.

### Final Answer
The number of solutions of the equation \( 2 \sin^3 x + 2 \cos^3 x - 3 \sin 2x + 2 = 0 \) in the interval \( [0, 4\pi] \) is **4**.