To solve the equation $\cos(2\theta) - \frac{1}{\sqrt{2}} = 0$, let's proceed step by step:

Step 1: Isolate $\cos(2\theta)$

$\cos(2\theta) = \frac{1}{\sqrt{2}}$

Step 2: Identify the angles where cosine equals $\frac{1}{\sqrt{2}}$

The cosine function equals $\frac{1}{\sqrt{2}}$ at angles: 2\theta = \frac{\pi}{4} + 2k\pi \quad \text{or} \quad 2\theta = \frac{7\pi}{4} + 2k\pi \quad \text{(where k is any integer)}.

Step 3: Solve for $\theta$

Divide through by 2 to isolate $\theta$: $\theta = \frac{\pi}{8} + k\pi \quad \text{or} \quad \theta = \frac{7\pi}{8} + k\pi.$

Step 4: Consider the Real Solutions

If no constraints are given (e.g., specific intervals), the real solutions are: \theta = \frac{\pi}{8} + k\pi \quad \text{and} \quad \theta = \frac{7\pi}{8} + k\pi, \quad \text{where k is any integer.}

Final Answer:

The general solution is: $\theta = \frac{\pi}{8} + k\pi \quad \text{or} \quad \theta = \frac{7\pi}{8} + k\pi, \quad k \in \mathbb{Z}.$

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Would you like further details or examples of how to apply this solution?

Related Questions:

1. How does the periodicity of cosine affect the solutions for trigonometric equations?
2. Can you explain the unit circle method to determine $\cos(x) = \frac{1}{\sqrt{2}}$?
3. What are the implications of limiting $\theta$ to a specific interval, like $[0, 2\pi)$?
4. How would this change if solving $\cos(2\theta) + \frac{1}{\sqrt{2}} = 0$?
5. How do solutions differ for $\sin(2\theta) = \frac{1}{\sqrt{2}}$ instead?

Tip:

Remember that cosine is periodic with period $2\pi$, so any solution can be adjusted by adding integer multiples of $2\pi$ to remain valid.