To solve the equation:

$\cos(2\theta) = \frac{\sqrt{3}}{2}$

we proceed as follows:

Step 1: Identify possible angles for $2\theta$

The cosine function equals $\frac{\sqrt{3}}{2}$ at two standard angles in the unit circle:

$2\theta = 30^\circ \quad \text{or} \quad 2\theta = 360^\circ - 30^\circ = 330^\circ$

In radians, this corresponds to:

$2\theta = \frac{\pi}{6} \quad \text{or} \quad 2\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$

Step 2: Solve for $\theta$

Now, divide both sides by 2 to find $\theta$:

1. From $2\theta = 30^\circ$, we get: $\theta = \frac{30^\circ}{2} = 15^\circ \quad \text{or} \quad \theta = \frac{\pi}{12} \text{ in radians}.$

2. From $2\theta = 330^\circ$, we get: $\theta = \frac{330^\circ}{2} = 165^\circ \quad \text{or} \quad \theta = \frac{11\pi}{12} \text{ in radians}.$

Step 3: General solution

The cosine function is periodic with period $360^\circ$ (or $2\pi$ radians), so the general solutions are:

$\theta = 15^\circ + 180^\circ n \quad \text{or} \quad \theta = 165^\circ + 180^\circ n$ where $n$ is any integer.

In radians, this becomes:

$\theta = \frac{\pi}{12} + n\pi \quad \text{or} \quad \theta = \frac{11\pi}{12} + n\pi$

Final Answer:

$\theta = 15^\circ + 180^\circ n \quad \text{or} \quad \theta = 165^\circ + 180^\circ n \quad \text{(degrees)}$ $\theta = \frac{\pi}{12} + n\pi \quad \text{or} \quad \theta = \frac{11\pi}{12} + n\pi \quad \text{(radians)}$

Would you like more details or explanations on any of these steps?

Related Questions:

1. What is the general form of solutions for cosine equations?
2. How do periodic functions affect solutions in trigonometry?
3. How do we convert between degrees and radians?
4. How does the unit circle help solve trigonometric equations?
5. How do you solve similar equations involving sine or tangent?

Tip: Always check for the periodicity of trigonometric functions when solving for angles.