Number of solution of the equation `sin^(2)x+(sin^(2)3x)/4=sinx sin 3x`, in `x epsilon(0,3pi)` equals

Options

A. `4`
B. `6`
C. `8`
D. `10`

Answer

To find the number of solutions for the equation

\[ \sin^2 x + \frac{\sin^2 3x}{4} = \sin x \sin 3x \]

in the interval \( x \in (0, 3\pi) \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:

\[ \sin^2 x + \frac{\sin^2 3x}{4} = \sin x \sin 3x \]

### Step 2: Rearranging the equation
We can rearrange the equation to:

\[ \sin^2 x - \sin x \sin 3x + \frac{\sin^2 3x}{4} = 0 \]

### Step 3: Completing the square
We can complete the square for the first two terms:

\[ \left(\sin x - \frac{1}{2} \sin 3x\right)^2 = 0 \]

This implies:

\[ \sin x - \frac{1}{2} \sin 3x = 0 \]

### Step 4: Express \(\sin 3x\) in terms of \(\sin x\)
Using the triple angle formula for sine, we have:

\[ \sin 3x = 3 \sin x - 4 \sin^3 x \]

Substituting this into our equation gives:

\[ \sin x - \frac{1}{2}(3 \sin x - 4 \sin^3 x) = 0 \]

### Step 5: Simplifying the equation
This simplifies to:

\[ \sin x - \frac{3}{2} \sin x + 2 \sin^3 x = 0 \]

Combining like terms results in:

\[ - \frac{1}{2} \sin x + 2 \sin^3 x = 0 \]

Factoring out \(\sin x\):

\[ \sin x \left(2 \sin^2 x - \frac{1}{2}\right) = 0 \]

### Step 6: Finding solutions
This gives us two cases to solve:

1. \(\sin x = 0\)
2. \(2 \sin^2 x - \frac{1}{2} = 0\)

#### Case 1: \(\sin x = 0\)
The solutions for \(\sin x = 0\) in the interval \( (0, 3\pi) \) are:

\[ x = n\pi \quad (n = 1, 2, 3) \]

Thus, we have solutions at \(x = \pi\), \(2\pi\).

#### Case 2: \(2 \sin^2 x - \frac{1}{2} = 0\)
Solving this gives:

\[ \sin^2 x = \frac{1}{4} \implies \sin x = \pm \frac{1}{2} \]

The solutions for \(\sin x = \frac{1}{2}\) are:

\[ x = \frac{\pi}{6} + 2n\pi \quad (n = 0, 1) \]
\[ x = \frac{5\pi}{6} + 2n\pi \quad (n = 0, 1) \]

The solutions for \(\sin x = -\frac{1}{2}\) are:

\[ x = \frac{7\pi}{6} + 2n\pi \quad (n = 0, 1) \]
\[ x = \frac{11\pi}{6} + 2n\pi \quad (n = 0, 1) \]

### Step 7: Counting the solutions
Now we count the valid solutions in the interval \( (0, 3\pi) \):

- From \(\sin x = 0\): \(x = \pi, 2\pi\) (2 solutions)
- From \(\sin x = \frac{1}{2}\): \(x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}\) (4 solutions)
- From \(\sin x = -\frac{1}{2}\): \(x = \frac{7\pi}{6}, \frac{11\pi}{6}\) (2 solutions)

### Total solutions
Adding these gives:

\[ 2 + 4 + 2 = 8 \]

Thus, the total number of solutions in the interval \( (0, 3\pi) \) is \(8\).

### Final Answer
The number of solutions of the equation in \( x \in (0, 3\pi) \) is **8**.