Find the values of ` theta` between `0^(@) and 360^(@)` that satisfy the equation
`sin ( theta + 30^(@)) = 2 cos (45^(@) + theta )`

Answer

To solve the equation \( \sin(\theta + 30^\circ) = 2 \cos(45^\circ + \theta) \) for \( \theta \) in the interval \( [0^\circ, 360^\circ] \), we will follow these steps:

### Step 1: Expand both sides using trigonometric identities
Using the sine and cosine addition formulas:
- \( \sin(a + b) = \sin a \cos b + \cos a \sin b \)
- \( \cos(a + b) = \cos a \cos b - \sin a \sin b \)

We can rewrite the left side:
\[ \sin(\theta + 30^\circ) = \sin \theta \cos 30^\circ + \cos \theta \sin 30^\circ \]
Substituting the values \( \cos 30^\circ = \frac{\sqrt{3}}{2} \) and \( \sin 30^\circ = \frac{1}{2} \):
\[ \sin(\theta + 30^\circ) = \sin \theta \cdot \frac{\sqrt{3}}{2} + \cos \theta \cdot \frac{1}{2} \]

Now for the right side:
\[ \cos(45^\circ + \theta) = \cos 45^\circ \cos \theta - \sin 45^\circ \sin \theta \]
Substituting the values \( \cos 45^\circ = \frac{1}{\sqrt{2}} \) and \( \sin 45^\circ = \frac{1}{\sqrt{2}} \):
\[ \cos(45^\circ + \theta) = \frac{1}{\sqrt{2}} \cos \theta - \frac{1}{\sqrt{2}} \sin \theta \]

Thus, the equation becomes:
\[ \frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta = 2 \left( \frac{1}{\sqrt{2}} \cos \theta - \frac{1}{\sqrt{2}} \sin \theta \right) \]

### Step 2: Simplify the equation
Expanding the right side:
\[ \frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta = \frac{2}{\sqrt{2}} \cos \theta - \frac{2}{\sqrt{2}} \sin \theta \]
This simplifies to:
\[ \frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta = \sqrt{2} \cos \theta - \sqrt{2} \sin \theta \]

### Step 3: Rearranging terms
Rearranging gives:
\[ \frac{\sqrt{3}}{2} \sin \theta + \sqrt{2} \sin \theta = \sqrt{2} \cos \theta - \frac{1}{2} \cos \theta \]
Factoring out \( \sin \theta \) and \( \cos \theta \):
\[ \left( \frac{\sqrt{3}}{2} + \sqrt{2} \right) \sin \theta = \left( \sqrt{2} - \frac{1}{2} \right) \cos \theta \]

### Step 4: Dividing both sides
Dividing both sides by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)):
\[ \tan \theta = \frac{\sqrt{2} - \frac{1}{2}}{\frac{\sqrt{3}}{2} + \sqrt{2}} \]

### Step 5: Solving for \( \theta \)
Let \( k = \frac{\sqrt{2} - \frac{1}{2}}{\frac{\sqrt{3}}{2} + \sqrt{2}} \). Thus:
\[ \theta = \tan^{-1}(k) \]

### Step 6: Finding all solutions in the interval
Since the tangent function is positive in the first and third quadrants, we have:
\[ \theta = \tan^{-1}(k) \quad \text{and} \quad \theta = 180^\circ + \tan^{-1}(k) \]

### Final Step: Calculate the values
Now, calculate \( k \) and find \( \theta \) values in the range \( [0^\circ, 360^\circ] \).