If ` 4 cos^2 x sin x - 2 sin^2 x = 3sinx ," then " x`=

Options

A. `n pi`
B. `n pi + (-1)^n pi/10`
C. `n pi + (-1)^n (3 pi)/10`
D. `n pi -(-1)^n (3pi)/10`

Answer

To solve the equation \( 4 \cos^2 x \sin x - 2 \sin^2 x = 3 \sin x \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 4 \cos^2 x \sin x - 2 \sin^2 x - 3 \sin x = 0 \]

### Step 2: Substitute \( \cos^2 x \)
Using the identity \( \cos^2 x = 1 - \sin^2 x \), we can rewrite the equation:
\[ 4 (1 - \sin^2 x) \sin x - 2 \sin^2 x - 3 \sin x = 0 \]
Expanding this gives:
\[ 4 \sin x - 4 \sin^3 x - 2 \sin^2 x - 3 \sin x = 0 \]
Combining like terms results in:
\[ -4 \sin^3 x - 2 \sin^2 x + \sin x = 0 \]

### Step 3: Factor out \( \sin x \)
Factoring out \( \sin x \) from the equation:
\[ \sin x (-4 \sin^2 x - 2 \sin x + 1) = 0 \]

### Step 4: Solve for \( \sin x = 0 \)
Setting \( \sin x = 0 \):
\[ x = n\pi, \quad n \in \mathbb{Z} \]

### Step 5: Solve the quadratic equation
Now, we solve the quadratic equation:
\[ -4 \sin^2 x - 2 \sin x + 1 = 0 \]
Multiplying through by -1 gives:
\[ 4 \sin^2 x + 2 \sin x - 1 = 0 \]
Using the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ \sin x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} \]
Calculating the discriminant:
\[ \sin x = \frac{-2 \pm \sqrt{4 + 16}}{8} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4} \]

### Step 6: Find the values of \( x \)
Now we have three cases to consider:
1. \( \sin x = 0 \) gives \( x = n\pi \)
2. \( \sin x = \frac{-1 + \sqrt{5}}{4} \)
3. \( \sin x = \frac{-1 - \sqrt{5}}{4} \)

For the second case, we can find \( x \) using the inverse sine function:
\[ x = \arcsin\left(\frac{-1 + \sqrt{5}}{4}\right) + 2k\pi \quad \text{and} \quad x = \pi - \arcsin\left(\frac{-1 + \sqrt{5}}{4}\right) + 2k\pi \]
where \( k \in \mathbb{Z} \).

For the third case, since \( \frac{-1 - \sqrt{5}}{4} \) is less than -1, it is not a valid sine value.

### Final Solution
Thus, the solutions for \( x \) are:
\[ x = n\pi, \quad n \in \mathbb{Z} \]
\[ x = \arcsin\left(\frac{-1 + \sqrt{5}}{4}\right) + 2k\pi \quad \text{and} \quad x = \pi - \arcsin\left(\frac{-1 + \sqrt{5}}{4}\right) + 2k\pi \]