Question

13. The general solution of the equation $\sin 2\theta = \frac{\sqrt{5}-1}{4}$ is

• A. $\theta = \frac{n\pi}{2} + (-1)^n \frac{\pi}{20}, n \in Z$
• B. $\theta = n\pi + (-1)^n \frac{\pi}{5}, n \in Z$
• C. $\theta = 2n\pi \pm \frac{\pi}{10}, n \in Z$
• D. $\theta = n\pi \pm \frac{\pi}{10}, n \in Z$

Answer 1

Answer: (A)

(a) $\theta = \frac{n\pi}{2} + (-1)^n \frac{\pi}{20}, n \in Z$

Answer 2

Given:

$$\sin 2\theta = \frac{\sqrt{5}-1}{4}$$

Recognize that

$$\frac{\sqrt{5}-1}{4} = \sin 18^\circ = \sin\left(\frac{\pi}{10}\right)$$

So,

$$\sin 2\theta = \sin\left(\frac{\pi}{10}\right)$$

This gives the solutions for $2\theta$:

$$2\theta = \frac{\pi}{10} + 2\pi k \quad \text{or} \quad 2\theta = \pi - \frac{\pi}{10} + 2\pi k, \quad k \in \mathbb{Z}$$

That is,

$$2\theta = \frac{\pi}{10} + 2\pi k \quad \text{or} \quad 2\theta = \frac{9\pi}{10} + 2\pi k$$

Dividing by 2:

$$\theta = \frac{\pi}{20} + \pi k \quad \text{or} \quad \theta = \frac{9\pi}{20} + \pi k, \quad k \in \mathbb{Z}$$

These can be combined into the compact form:

$$\theta = \frac{n\pi}{2} + (-1)^n \frac{\pi}{20}, \quad n \in \mathbb{Z}$$

which is equivalent to option (a).