Find all the solution of ` 4 cos^(2)x sinx -2 sin^(2) x=3 sinx `

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
Start by rewriting \( \cos^2 x \) in terms of \( \sin x \):
\[ \cos^2 x = 1 - \sin^2 x \]
Substituting this into the equation gives:
\[ 4(1 - \sin^2 x) \sin x - 2 \sin^2 x = 3 \sin x \]

### Step 2: Expand and Rearrange
Now, expand the left-hand side:
\[ 4 \sin x - 4 \sin^3 x - 2 \sin^2 x = 3 \sin x \]
Rearranging this, we get:
\[ -4 \sin^3 x - 2 \sin^2 x + 4 \sin x - 3 \sin x = 0 \]
This simplifies to:
\[ -4 \sin^3 x - 2 \sin^2 x + \sin x = 0 \]

### Step 3: Factor Out \( \sin x \)
Factor out \( \sin x \):
\[ \sin x (-4 \sin^2 x - 2 \sin x + 1) = 0 \]
This gives us two cases to solve:
1. \( \sin x = 0 \)
2. \( -4 \sin^2 x - 2 \sin x + 1 = 0 \)

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

### Step 5: Solve the Quadratic Equation
Now, 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 \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 4, b = 2, c = -1 \)
\[ \sin x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} \]
\[ = \frac{-2 \pm \sqrt{4 + 16}}{8} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4} \]

### Step 6: Find Values of \( x \)
Now we have:
\[ \sin x = \frac{-1 + \sqrt{5}}{4} \quad \text{and} \quad \sin x = \frac{-1 - \sqrt{5}}{4} \]
The second value \( \frac{-1 - \sqrt{5}}{4} \) is less than -1, which is not possible for sine. So we only consider:
\[ \sin x = \frac{-1 + \sqrt{5}}{4} \]

### Step 7: General Solutions
The general solutions for \( \sin x = k \) are given by:
\[ x = n\pi + (-1)^n \arcsin\left(\frac{-1 + \sqrt{5}}{4}\right), \quad n \in \mathbb{Z} \]

### Final Solution
Combining both cases, the complete solution set is:
\[ x = n\pi \quad \text{or} \quad x = n\pi + (-1)^n \arcsin\left(\frac{-1 + \sqrt{5}}{4}\right), \quad n \in \mathbb{Z} \]