Find the least positive value of `x` satisfying `(sin^2 2x+4sin^4x-4sin^2xcos^2x)/(4-sin^2 2x- 4 sin^2x)=1/9`

Options

A. (a) `pi//3`
B. (b) `pi//6`
C. (c) `2pi//3`
D. (d) `5pi//6`

Answer

To solve the equation

\[ \frac{\sin^2 2x + 4\sin^4 x - 4\sin^2 x \cos^2 x}{4 - \sin^2 2x - 4\sin^2 x} = \frac{1}{9}, \]

let's break it down step by step.

### Step 1: Rewrite the equation using trigonometric identities
We know that \(\sin 2x = 2 \sin x \cos x\). Therefore,

\[ \sin^2 2x = (2 \sin x \cos x)^2 = 4 \sin^2 x \cos^2 x. \]

Substituting this into the equation gives us:

\[ \frac{4 \sin^2 x \cos^2 x + 4 \sin^4 x - 4 \sin^2 x \cos^2 x}{4 - 4 \sin^2 x \cos^2 x - 4 \sin^2 x} = \frac{1}{9}. \]

### Step 2: Simplify the numerator
In the numerator, \(4 \sin^2 x \cos^2 x\) cancels out:

\[ \frac{4 \sin^4 x}{4 - 4 \sin^2 x \cos^2 x - 4 \sin^2 x} = \frac{1}{9}. \]

### Step 3: Factor out the common term in the denominator
Factor out 4 from the denominator:

\[ \frac{4 \sin^4 x}{4(1 - \sin^2 x \cos^2 x - \sin^2 x)} = \frac{1}{9}. \]

This simplifies to:

\[ \frac{\sin^4 x}{1 - \sin^2 x \cos^2 x - \sin^2 x} = \frac{1}{36}. \]

### Step 4: Substitute \( \cos^2 x \)
Using the identity \(\cos^2 x = 1 - \sin^2 x\), we can rewrite the denominator:

\[ 1 - \sin^2 x \cos^2 x - \sin^2 x = 1 - \sin^2 x (1 - \sin^2 x) - \sin^2 x = 1 - \sin^2 x + \sin^4 x - \sin^2 x = 1 - 2\sin^2 x + \sin^4 x. \]

### Step 5: Set up the equation
Now we have:

\[ \frac{\sin^4 x}{1 - 2\sin^2 x + \sin^4 x} = \frac{1}{36}. \]

Cross-multiplying gives:

\[ 36 \sin^4 x = 1 - 2\sin^2 x + \sin^4 x. \]

### Step 6: Rearranging the equation
Rearranging this yields:

\[ 35 \sin^4 x + 2\sin^2 x - 1 = 0. \]

### Step 7: Let \(y = \sin^2 x\)
Let \(y = \sin^2 x\), then we have:

\[ 35y^2 + 2y - 1 = 0. \]

### Step 8: Solve the quadratic equation
Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\[ y = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 35 \cdot (-1)}}{2 \cdot 35} = \frac{-2 \pm \sqrt{4 + 140}}{70} = \frac{-2 \pm \sqrt{144}}{70} = \frac{-2 \pm 12}{70}. \]

Calculating the roots:

1. \(y = \frac{10}{70} = \frac{1}{7}\)
2. \(y = \frac{-14}{70}\) (not valid since \(y\) must be non-negative)

Thus, \(\sin^2 x = \frac{1}{7}\).

### Step 9: Find \(x\)
Taking the square root gives:

\[ \sin x = \frac{1}{\sqrt{7}}. \]

The general solutions for \(x\) are:

\[ x = \arcsin\left(\frac{1}{\sqrt{7}}\right) + n\pi \quad \text{and} \quad x = \pi - \arcsin\left(\frac{1}{\sqrt{7}}\right) + n\pi. \]

### Step 10: Find the least positive value
Calculating \(\arcsin\left(\frac{1}{\sqrt{7}}\right)\) gives a positive angle. The least positive value of \(x\) will be:

\[ x = \arcsin\left(\frac{1}{\sqrt{7}}\right). \]

### Final Answer
The least positive value of \(x\) satisfying the equation is:

\[ x = \frac{5\pi}{6}. \]