Solve: `2sin^(2)theta+sin^(2)2theta=2`

Options

A. `(npi-pi/3)`
B. `npi-pi/6`
C. `npi-pi/4`
D. `npi-pi/12`

Answer

To solve the equation \( 2\sin^2\theta + \sin^2(2\theta) = 2 \), we can follow these steps:

### Step 1: Rewrite the equation
The given equation is:
\[ 2\sin^2\theta + \sin^2(2\theta) = 2 \]

### Step 2: Use the double angle identity
We know that \( \sin(2\theta) = 2\sin\theta\cos\theta \). Therefore, we can express \( \sin^2(2\theta) \) as:
\[ \sin^2(2\theta) = (2\sin\theta\cos\theta)^2 = 4\sin^2\theta\cos^2\theta \]

### Step 3: Substitute into the equation
Substituting this back into the equation gives:
\[ 2\sin^2\theta + 4\sin^2\theta\cos^2\theta = 2 \]

### Step 4: Factor out common terms
Let \( x = \sin^2\theta \). Then, we can rewrite the equation as:
\[ 2x + 4x(1 - x) = 2 \]
This simplifies to:
\[ 2x + 4x - 4x^2 = 2 \]
or:
\[ 6x - 4x^2 = 2 \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ 4x^2 - 6x + 2 = 0 \]

### Step 6: Solve the quadratic equation
Now we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 4, b = -6, c = 2 \):
\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 4 \cdot 2}}{2 \cdot 4} \]
\[ x = \frac{6 \pm \sqrt{36 - 32}}{8} \]
\[ x = \frac{6 \pm \sqrt{4}}{8} \]
\[ x = \frac{6 \pm 2}{8} \]
This gives us two solutions:
\[ x_1 = \frac{8}{8} = 1 \quad \text{and} \quad x_2 = \frac{4}{8} = \frac{1}{2} \]

### Step 7: Back substitute for \( \sin^2\theta \)
Thus, we have:
1. \( \sin^2\theta = 1 \) which implies \( \sin\theta = \pm 1 \)
2. \( \sin^2\theta = \frac{1}{2} \) which implies \( \sin\theta = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \)

### Step 8: Find the angles
For \( \sin\theta = 1 \):
\[ \theta = \frac{\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]

For \( \sin\theta = -1 \):
\[ \theta = \frac{3\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]

For \( \sin\theta = \frac{\sqrt{2}}{2} \):
\[ \theta = \frac{\pi}{4} + 2n\pi \quad \text{or} \quad \theta = \frac{3\pi}{4} + 2n\pi \quad (n \in \mathbb{Z}) \]

For \( \sin\theta = -\frac{\sqrt{2}}{2} \):
\[ \theta = \frac{5\pi}{4} + 2n\pi \quad \text{or} \quad \theta = \frac{7\pi}{4} + 2n\pi \quad (n \in \mathbb{Z}) \]

### Final Solution
Thus, the complete solution set is:
\[ \theta = \frac{\pi}{2} + 2n\pi, \quad \theta = \frac{3\pi}{2} + 2n\pi, \quad \theta = \frac{\pi}{4} + 2n\pi, \quad \theta = \frac{3\pi}{4} + 2n\pi, \quad \theta = \frac{5\pi}{4} + 2n\pi, \quad \theta = \frac{7\pi}{4} + 2n\pi \quad (n \in \mathbb{Z}) \]