If `1+2sin x+3sin^(2)x+4sin^(3)x+...` upto infinite terms = 4 and number of solutions of the equation in `[ (-3pi)/(2), 4pi] ` is k.
The value of `|(cos2x-1)/(sin2x)|` is equal to

Options

A. 1
B. `sqrt(3)`
C. `2-sqrt(3)`
D. `(1)/(sqrt(3))`

Answer

To solve the problem step by step, we will follow the same logic as in the video transcript.

### Step 1: Set up the equation
We are given the series:
\[ 1 + 2\sin x + 3\sin^2 x + 4\sin^3 x + \ldots = 4 \]
Let \( s = \sin x \). Thus, we can rewrite the equation as:
\[ 1 + 2s + 3s^2 + 4s^3 + \ldots = 4 \]

### Step 2: Recognize the series
The left-hand side can be expressed as:
\[ \sum_{n=1}^{\infty} n s^{n-1} = \frac{1}{(1-s)^2} \]
This is the formula for the sum of an infinite series where the first term is 1 and the common ratio is \( s \).

### Step 3: Set up the equation
Now we can set up the equation:
\[ 4 = \frac{1}{(1-s)^2} \]

### Step 4: Cross-multiply and simplify
Cross-multiplying gives:
\[ 4(1 - s)^2 = 1 \]
Expanding this:
\[ 4(1 - 2s + s^2) = 1 \]
\[ 4 - 8s + 4s^2 = 1 \]
Rearranging gives:
\[ 4s^2 - 8s + 3 = 0 \]

### Step 5: Solve the quadratic equation
Now we can use the quadratic formula \( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 4, b = -8, c = 3 \):
\[ s = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 4 \cdot 3}}{2 \cdot 4} \]
\[ s = \frac{8 \pm \sqrt{64 - 48}}{8} \]
\[ s = \frac{8 \pm \sqrt{16}}{8} \]
\[ s = \frac{8 \pm 4}{8} \]

This gives us two possible solutions:
1. \( s = \frac{12}{8} = \frac{3}{2} \) (not valid since \( s \) must be in \([-1, 1]\))
2. \( s = \frac{4}{8} = \frac{1}{2} \)

### Step 6: Find the number of solutions for \( \sin x = \frac{1}{2} \)
Now, we need to find the solutions for \( \sin x = \frac{1}{2} \) in the interval \( \left[-\frac{3\pi}{2}, 4\pi\right] \).

The general solutions for \( \sin x = \frac{1}{2} \) are:
\[ x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + 2k\pi \]
for \( k \in \mathbb{Z} \).

### Step 7: Determine the values of \( k \)
We will find the integer values of \( k \) such that \( x \) lies within the specified interval.

1. For \( x = \frac{\pi}{6} + 2k\pi \):
- For \( k = -1 \): \( x = \frac{\pi}{6} - 2\pi = \frac{\pi - 12\pi}{6} = -\frac{11\pi}{6} \) (valid)
- For \( k = 0 \): \( x = \frac{\pi}{6} \) (valid)
- For \( k = 1 \): \( x = \frac{\pi}{6} + 2\pi = \frac{\pi + 12\pi}{6} = \frac{13\pi}{6} \) (valid)
- For \( k = 2 \): \( x = \frac{\pi}{6} + 4\pi = \frac{\pi + 24\pi}{6} = \frac{25\pi}{6} \) (not valid since it exceeds \( 4\pi \))

2. For \( x = \frac{5\pi}{6} + 2k\pi \):
- For \( k = -1 \): \( x = \frac{5\pi}{6} - 2\pi = \frac{5\pi - 12\pi}{6} = -\frac{7\pi}{6} \) (valid)
- For \( k = 0 \): \( x = \frac{5\pi}{6} \) (valid)
- For \( k = 1 \): \( x = \frac{5\pi}{6} + 2\pi = \frac{5\pi + 12\pi}{6} = \frac{17\pi}{6} \) (valid)
- For \( k = 2 \): \( x = \frac{5\pi}{6} + 4\pi = \frac{5\pi + 24\pi}{6} = \frac{29\pi}{6} \) (not valid since it exceeds \( 4\pi \))

### Step 8: Count the valid solutions
The valid solutions are:
- From \( \sin x = \frac{1}{2} \):
- \( -\frac{11\pi}{6} \)
- \( \frac{\pi}{6} \)
- \( \frac{13\pi}{6} \)
- From \( \sin x = \frac{1}{2} \):
- \( -\frac{7\pi}{6} \)
- \( \frac{5\pi}{6} \)
- \( \frac{17\pi}{6} \)

Total valid solutions = 6.

### Step 9: Calculate \( \left| \frac{\cos 2x - 1}{\sin 2x} \right| \)
Now we need to calculate:
\[ \left| \frac{\cos 2x - 1}{\sin 2x} \right| \]

Using the double angle formulas:
- \( \cos 2x = \cos^2 x - \sin^2 x \)
- \( \sin 2x = 2 \sin x \cos x \)

Substituting \( \sin x = \frac{1}{2} \) and \( \cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \):

Now substituting these values:
\[ \cos 2x = \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \]
Thus,
\[ \cos 2x - 1 = \frac{1}{2} - 1 = -\frac{1}{2} \]

And for \( \sin 2x \):
\[ \sin 2x = 2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \]

Now substituting into the expression:
\[ \left| \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right| = \left| -\frac{1}{\sqrt{3}} \right| = \frac{1}{\sqrt{3}} \]

### Final Answer
Thus, the value of \( \left| \frac{\cos 2x - 1}{\sin 2x} \right| \) is:
\[ \frac{1}{\sqrt{3}} \]