The number of solutions of the equation ` sin^(4) theta - 2 sin^(2) theta +1 = 0 ` which lie between 0 and ` 2 pi ` is :

Options

A. 0
B. 2
C. 4
D. 8

Answer

To find the number of solutions of the equation \( \sin^4 \theta - 2 \sin^2 \theta + 1 = 0 \) that lie between \( 0 \) and \( 2\pi \), we can follow these steps:

### Step 1: Substitute \( t \) for \( \sin^2 \theta \)
Let \( t = \sin^2 \theta \). Then the equation becomes:
\[ t^2 - 2t + 1 = 0 \]

### Step 2: Factor the quadratic equation
The equation can be factored as:
\[ (t - 1)^2 = 0 \]

### Step 3: Solve for \( t \)
Setting the factor equal to zero gives:
\[ t - 1 = 0 \implies t = 1 \]

### Step 4: Substitute back to find \( \sin^2 \theta \)
Since \( t = \sin^2 \theta \), we have:
\[ \sin^2 \theta = 1 \]

### Step 5: Solve for \( \sin \theta \)
Taking the square root of both sides, we find:
\[ \sin \theta = \pm 1 \]

### Step 6: Find the angles \( \theta \)
The values of \( \theta \) that satisfy \( \sin \theta = 1 \) and \( \sin \theta = -1 \) in the interval \( [0, 2\pi] \) are:
- For \( \sin \theta = 1 \):
\[ \theta = \frac{\pi}{2} \]
- For \( \sin \theta = -1 \):
\[ \theta = \frac{3\pi}{2} \]

### Step 7: Count the solutions
Thus, the solutions in the interval \( [0, 2\pi] \) are:
- \( \theta = \frac{\pi}{2} \)
- \( \theta = \frac{3\pi}{2} \)

Therefore, there are a total of **2 solutions**.

### Final Answer
The number of solutions of the equation \( \sin^4 \theta - 2 \sin^2 \theta + 1 = 0 \) which lie between \( 0 \) and \( 2\pi \) is **2**.

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