If `sin2theta = cos3theta`, where `0 le theta le (pi)/(2)`, then what is sin`theta` equal to ?

Options

A. `(sqrt3+1)/(4)`
B. `(sqrt5-1)/(4)`
C. `(sqrt5+1)/(6)`
D. `(sqrt5-1)/(16)`

Answer

To solve the equation \( \sin 2\theta = \cos 3\theta \) for \( \theta \) in the range \( 0 \leq \theta \leq \frac{\pi}{2} \), we will follow these steps:

### Step 1: Use the double angle and triple angle identities
We know that:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
and
\[ \cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta \]
Thus, we can rewrite the equation as:
\[ 2 \sin \theta \cos \theta = 4 \cos^3 \theta - 3 \cos \theta \]

### Step 2: Rearrange the equation
We can rearrange the equation to bring all terms to one side:
\[ 2 \sin \theta \cos \theta - 4 \cos^3 \theta + 3 \cos \theta = 0 \]

### Step 3: Factor out common terms
Notice that we can factor out \( \cos \theta \):
\[ \cos \theta (2 \sin \theta - 4 \cos^2 \theta + 3) = 0 \]
This gives us two cases to consider:
1. \( \cos \theta = 0 \)
2. \( 2 \sin \theta - 4 \cos^2 \theta + 3 = 0 \)

### Step 4: Analyze the first case
For \( \cos \theta = 0 \):
- This occurs at \( \theta = \frac{\pi}{2} \).
- We need to check if this satisfies the original equation:
\[ \sin 2\left(\frac{\pi}{2}\right) = \sin \pi = 0 \]
\[ \cos 3\left(\frac{\pi}{2}\right) = \cos \frac{3\pi}{2} = 0 \]
Both sides are equal, so \( \theta = \frac{\pi}{2} \) is a valid solution.

### Step 5: Analyze the second case
Now, we solve:
\[ 2 \sin \theta - 4 \cos^2 \theta + 3 = 0 \]
Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \):
\[ 2 \sin \theta - 4(1 - \sin^2 \theta) + 3 = 0 \]
Expanding this gives:
\[ 2 \sin \theta - 4 + 4 \sin^2 \theta + 3 = 0 \]
This simplifies to:
\[ 4 \sin^2 \theta + 2 \sin \theta - 1 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 4, b = 2, c = -1 \):
\[ \sin \theta = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} \]
\[ = \frac{-2 \pm \sqrt{4 + 16}}{8} \]
\[ = \frac{-2 \pm \sqrt{20}}{8} \]
\[ = \frac{-2 \pm 2\sqrt{5}}{8} \]
\[ = \frac{-1 \pm \sqrt{5}}{4} \]

### Step 7: Determine valid solutions
Since \( \theta \) is in the first quadrant, \( \sin \theta \) must be positive:
\[ \sin \theta = \frac{-1 + \sqrt{5}}{4} \]
is the valid solution, while \( \frac{-1 - \sqrt{5}}{4} \) is negative and thus not valid.

### Final Answer
Thus, the solution for \( \sin \theta \) is:
\[ \sin \theta = \frac{-1 + \sqrt{5}}{4} \]

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