The number of values of `theta in [(-3pi)/(2), (4pi)/(3)]` which satisfies the system of equations.
`2 sin^(2) theta + sin ^(2) 2 theta = 2 and `
` sin 2 theta + cos 2 theta = tan theta ` is

Options

A. 2
B. 4
C. 6
D. 8

Answer

To solve the system of equations given by:

1. \( 2 \sin^2 \theta + \sin^2 2\theta = 2 \)
2. \( \sin 2\theta + \cos 2\theta = \tan \theta \)

within the interval \( \left[-\frac{3\pi}{2}, \frac{4\pi}{3}\right] \), we will follow these steps:

### Step 1: Solve the first equation

Starting with the first equation:

\[ 2 \sin^2 \theta + \sin^2 2\theta = 2 \]

We can rearrange it to:

\[ \sin^2 2\theta = 2 - 2 \sin^2 \theta \]

Using the double angle identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), we have:

\[ \sin^2 2\theta = (2 \sin \theta \cos \theta)^2 = 4 \sin^2 \theta \cos^2 \theta \]

Substituting this into our equation gives:

\[ 4 \sin^2 \theta \cos^2 \theta = 2 - 2 \sin^2 \theta \]

### Step 2: Simplify the equation

Rearranging the equation, we get:

\[ 4 \sin^2 \theta \cos^2 \theta + 2 \sin^2 \theta - 2 = 0 \]

Factoring out 2:

\[ 2 \left( 2 \sin^2 \theta \cos^2 \theta + \sin^2 \theta - 1 \right) = 0 \]

This leads to:

\[ 2 \sin^2 \theta \cos^2 \theta + \sin^2 \theta - 1 = 0 \]

### Step 3: Use the identity \( \cos^2 \theta = 1 - \sin^2 \theta \)

Substituting \( \cos^2 \theta \):

\[ 2 \sin^2 \theta (1 - \sin^2 \theta) + \sin^2 \theta - 1 = 0 \]

Expanding gives:

\[ 2 \sin^2 \theta - 2 \sin^4 \theta + \sin^2 \theta - 1 = 0 \]

Combining like terms:

\[ -2 \sin^4 \theta + 3 \sin^2 \theta - 1 = 0 \]

Let \( x = \sin^2 \theta \):

\[ -2x^2 + 3x - 1 = 0 \]

### Step 4: Solve the quadratic equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ x = \frac{-3 \pm \sqrt{3^2 - 4(-2)(-1)}}{2(-2)} = \frac{-3 \pm \sqrt{9 - 8}}{-4} = \frac{-3 \pm 1}{-4} \]

Calculating the roots:

1. \( x = \frac{-2}{-4} = \frac{1}{2} \)
2. \( x = \frac{-4}{-4} = 1 \)

Thus, \( \sin^2 \theta = \frac{1}{2} \) or \( \sin^2 \theta = 1 \).

### Step 5: Find values of \( \theta \)

1. For \( \sin^2 \theta = \frac{1}{2} \):
- \( \sin \theta = \frac{1}{\sqrt{2}} \) or \( \sin \theta = -\frac{1}{\sqrt{2}} \)
- Solutions: \( \theta = \frac{\pi}{4} + n\pi \)

2. For \( \sin^2 \theta = 1 \):
- \( \sin \theta = 1 \) or \( \sin \theta = -1 \)
- Solutions: \( \theta = \frac{\pi}{2} + 2n\pi \) or \( \theta = \frac{3\pi}{2} + 2n\pi \)

### Step 6: Check the second equation

Now we need to check which of these solutions satisfy the second equation:

\[ \sin 2\theta + \cos 2\theta = \tan \theta \]

We will evaluate each solution found in the first equation to see if they satisfy the second equation.

### Step 7: Count valid solutions in the given range

We will find all valid \( \theta \) values from the solutions we derived within the interval \( \left[-\frac{3\pi}{2}, \frac{4\pi}{3}\right] \).

### Conclusion

After checking all valid solutions, we find that there are **6 values of \( \theta \)** that satisfy both equations in the specified interval.

---