Find number of solution of the equation `2 sin x+5 sin^(2) x+8sin^(3)x+... oo=1` for `x in [0, 2pi]`.

Explanation

To solve the equation \( 2 \sin x + 5 \sin^2 x + 8 \sin^3 x + \ldots = 1 \) for \( x \) in the interval \( [0, 2\pi] \), we can follow these steps:

### Step 1: Recognize the series
The left-hand side of the equation is an infinite series. We can express it as:
\[ S = 2 \sin x + 5 \sin^2 x + 8 \sin^3 x + \ldots \]
This series can be rewritten in a more manageable form.

### Step 2: Multiply both sides by \( \sin x \)
Multiply the entire equation by \( \sin x \):
\[ 2 \sin^2 x + 5 \sin^3 x + 8 \sin^4 x + \ldots = \sin x \]
Let’s denote this new equation as Equation (1).

### Step 3: Subtract the original series from the new equation
Now, we can subtract the original series from this new equation:
\[ (2 \sin^2 x + 5 \sin^3 x + 8 \sin^4 x + \ldots) - (2 \sin x + 5 \sin^2 x + 8 \sin^3 x + \ldots) = \sin x - 1 \]
This simplifies to:
\[ (2 \sin^2 x - 2 \sin x) + (5 \sin^3 x - 5 \sin^2 x) + (8 \sin^4 x - 8 \sin^3 x) + \ldots = \sin x - 1 \]

### Step 4: Simplify the equation
Rearranging gives us:
\[ 2 \sin^2 x + 3 \sin^3 x + 3 \sin^4 x + \ldots = 1 - \sin x \]
Let’s denote this new equation as Equation (2).

### Step 5: Factor out common terms
We can factor out \( 3 \sin x \) from the left-hand side:
\[ 3 \sin x (1 + \sin x + \sin^2 x + \ldots) = 1 - \sin x \]
The series \( 1 + \sin x + \sin^2 x + \ldots \) is a geometric series with first term \( 1 \) and common ratio \( \sin x \).

### Step 6: Sum the geometric series
The sum of an infinite geometric series is given by:
\[ \text{Sum} = \frac{A}{1 - R} = \frac{1}{1 - \sin x} \]
Thus, we have:
\[ 3 \sin x \cdot \frac{1}{1 - \sin x} = 1 - \sin x \]

### Step 7: Rearranging the equation
Multiplying through by \( 1 - \sin x \) gives:
\[ 3 \sin x = (1 - \sin x)(1 - \sin x) \]
Expanding the right-hand side:
\[ 3 \sin x = 1 - 2 \sin x + \sin^2 x \]
Rearranging gives:
\[ \sin^2 x - 5 \sin x + 1 = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ \sin x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ \sin x = \frac{5 \pm \sqrt{25 - 4}}{2} \]
\[ \sin x = \frac{5 \pm \sqrt{21}}{2} \]

### Step 9: Determine valid solutions
Now we need to check if these values of \( \sin x \) lie within the range \([-1, 1]\):
1. \( \sin x = \frac{5 + \sqrt{21}}{2} \) is greater than 1 (not valid).
2. \( \sin x = \frac{5 - \sqrt{21}}{2} \) needs to be checked.

Calculating \( \frac{5 - \sqrt{21}}{2} \):
- Since \( \sqrt{21} \approx 4.58 \), we find \( 5 - \sqrt{21} \approx 0.42 \).
- Thus, \( \frac{5 - \sqrt{21}}{2} \approx 0.21 \) (valid).

### Step 10: Find the number of solutions in \( [0, 2\pi] \)
The value \( \sin x = \frac{5 - \sqrt{21}}{2} \) corresponds to two angles in the interval \( [0, 2\pi] \):
1. \( x = \arcsin\left(\frac{5 - \sqrt{21}}{2}\right) \)
2. \( x = \pi - \arcsin\left(\frac{5 - \sqrt{21}}{2}\right) \)

Thus, the total number of solutions is **2**.