Number of solutions of the equation `2 sin^3x + 6 sin^2x -sin x-3=0` in `(0, 2pi),` are

Answer

To find the number of solutions of the equation \(2 \sin^3 x + 6 \sin^2 x - \sin x - 3 = 0\) in the interval \((0, 2\pi)\), we can follow these steps:

### Step 1: Substitute \(t = \sin x\)

We rewrite the equation in terms of \(t\):
\[ 2t^3 + 6t^2 - t - 3 = 0 \]

### Step 2: Factor the polynomial

To factor the polynomial, we can use the Rational Root Theorem or synthetic division. We can check for possible rational roots. Testing \(t = -3\):
\[ 2(-3)^3 + 6(-3)^2 - (-3) - 3 = 2(-27) + 6(9) + 3 - 3 = -54 + 54 + 3 - 3 = 0 \]
Thus, \(t = -3\) is a root.

### Step 3: Perform polynomial long division

Now we divide \(2t^3 + 6t^2 - t - 3\) by \(t + 3\):
1. Divide \(2t^3\) by \(t\) to get \(2t^2\).
2. Multiply \(2t^2\) by \(t + 3\) to get \(2t^3 + 6t^2\).
3. Subtract:
\[ (2t^3 + 6t^2 - t - 3) - (2t^3 + 6t^2) = -t - 3 \]
4. Divide \(-t\) by \(t\) to get \(-1\).
5. Multiply \(-1\) by \(t + 3\) to get \(-t - 3\).
6. Subtract:
\[ (-t - 3) - (-t - 3) = 0 \]

Thus, we can write:
\[ 2t^3 + 6t^2 - t - 3 = (t + 3)(2t^2 - 1) \]

### Step 4: Solve the quadratic equation

Now we solve:
\[ 2t^2 - 1 = 0 \]
This gives:
\[ 2t^2 = 1 \quad \Rightarrow \quad t^2 = \frac{1}{2} \quad \Rightarrow \quad t = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \]

### Step 5: Identify valid solutions for \(\sin x\)

We have the following potential values for \(\sin x\):
1. \(t = -3\) (not valid since \(\sin x\) must be in \([-1, 1]\))
2. \(t = \frac{\sqrt{2}}{2}\)
3. \(t = -\frac{\sqrt{2}}{2}\)

### Step 6: Find the angles corresponding to \(\sin x\)

1. For \(t = \frac{\sqrt{2}}{2}\):
- \(x = \frac{\pi}{4}\)
- \(x = \frac{3\pi}{4}\)

2. For \(t = -\frac{\sqrt{2}}{2}\):
- \(x = \frac{5\pi}{4}\)
- \(x = \frac{7\pi}{4}\)

### Step 7: Count the solutions in the interval \((0, 2\pi)\)

The solutions in the interval \((0, 2\pi)\) are:
1. \(x = \frac{\pi}{4}\)
2. \(x = \frac{3\pi}{4}\)
3. \(x = \frac{5\pi}{4}\)
4. \(x = \frac{7\pi}{4}\)

Thus, the total number of solutions is **4**.