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We are tasked with solving the trigonometric equation \( 2\sin(2\theta) = \cos(2\theta + 30^\circ) \) for \( 0 \leq \theta < 180^\circ \), giving the answer to one decimal place.

Let's break down the steps to solve it.

### Step 1: Use trigonometric identities
We can expand the right-hand side of the equation using the cosine angle addition formula:
\[
\cos(2\theta + 30^\circ) = \cos(2\theta)\cos(30^\circ) - \sin(2\theta)\sin(30^\circ)
\]
where \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \) and \( \sin(30^\circ) = \frac{1}{2} \).

Substituting these values into the equation:
\[
2\sin(2\theta) = \cos(2\theta) \cdot \frac{\sqrt{3}}{2} - \sin(2\theta) \cdot \frac{1}{2}
\]

### Step 2: Simplify the equation
Multiply both sides of the equation by 2 to eliminate the fractions:
\[
4\sin(2\theta) = \sqrt{3} \cos(2\theta) - \sin(2\theta)
\]

Move all terms involving \( \sin(2\theta) \) to one side of the equation:
\[
4\sin(2\theta) + \sin(2\theta) = \sqrt{3} \cos(2\theta)
\]
\[
5\sin(2\theta) = \sqrt{3} \cos(2\theta)
\]

### Step 3: Divide by \( \cos(2\theta) \) (assuming \( \cos(2\theta) \neq 0 \))
\[
\frac{\sin(2\theta)}{\cos(2\theta)} = \frac{\sqrt{3}}{5}
\]
This simplifies to:
\[
\tan(2\theta) = \frac{\sqrt{3}}{5}
\]

### Step 4: Solve for \( 2\theta \)
Take the inverse tangent (arctan) of both sides:
\[
2\theta = \tan^{-1}\left( \frac{\sqrt{3}}{5} \right)
\]

Using a calculator to find the angle:
\[
2\theta \approx 19.1^\circ
\]

### Step 5: Solve for \( \theta \)
Now, divide by 2 to get \( \theta \):
\[
\theta \approx \frac{19.1^\circ}{2} = 9.55^\circ
\]

### Step 6: Check for additional solutions
Since \( \tan(2\theta) \) is periodic with a period of \( 180^\circ \), additional solutions may be found by adding multiples of \( 180^\circ \) to \( 2\theta \):
\[
2\theta = 19.1^\circ + 180^\circ n
\]
for integers \( n \).

For \( n = 0 \), we already found \( \theta \approx 9.6^\circ \).

For \( n = 1 \):
\[
2\theta = 19.1^\circ + 180^\circ = 199.1^\circ
\]
\[
\theta = \frac{199.1^\circ}{2} = 99.6^\circ
\]

So, the two solutions in the range \( 0^\circ \leq \theta < 180^\circ \) are:
\[
\theta \approx 9.6^\circ \quad \text{and} \quad \theta \approx 99.6^\circ
\]

### Final Answer:
The solutions are \( \theta = 9.6^\circ \) and \( \theta = 99.6^\circ \).

Would you like further details or clarifications on any step? 

Here are 5 related questions:
1. What is the cosine angle addition formula?
2. How do you solve equations involving \( \sin \) and \( \cos \) terms?
3. What is the period of the tangent function?
4. Why does adding multiples of 180 degrees yield additional solutions in trigonometric equations?
5. How do you compute inverse trigonometric functions using a calculator?

**Tip:** Always check the range of your solutions when solving trigonometric equations, as trigonometric functions are periodic.

Solve the equation 2sin(2θ) = cos(2θ + 30°) for 0 ≤ θ < 180°, giving your answers to one decimal place.

Learn how to solve the trigonometric equation 2sin(2θ) = cos(2θ + 30°) for angles within the range 0° ≤ θ < 180°, with a detailed step-by-step solution. This problem covers key trigonometric concepts such as the cosine addition formula, tangent function, and inverse trigonometric calculations. Suitable for students in Grades 10-12.

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