What are the solutions of \( \cos 4 \theta=\cos 2 \theta=\sin 30 \) ? Seleet It thater(s) \( \frac{7 \pi}{6}+2 \pi \pi \) \( \frac{n \pi}{4} \) \( \frac{7 \pi}{6}+n \pi \) \( \frac{\pi}{4}+n \pi \) \( 0+\frac{n \pi}{3} \) \( \frac{11 \pi}{6}+m \pi \) \( \frac{3 \pi}{4}+n \pi \) \( \frac{11 \pi}{6}+2 n \pi \)

To solve the equation \( \cos 4\theta = \cos 2\theta = \sin 30^\circ \), we first need to evaluate \( \sin 30^\circ \): \[ \sin 30^\circ = \frac{1}{2} \] Now we can rewrite the equations as: 1. \( \cos 4\theta = \frac{1}{2} \) 2. \( \cos 2\theta = \frac{1}{2} \) ### Step 1: Solve \( \cos 4\theta = \frac{1}{2} \) The general solutions for \( \cos x = \frac{1}{2} \) are: \[ x = 2n\pi \pm \frac{\pi}{3} \quad (n \in \mathbb{Z}) \] Substituting \( x = 4\theta \): \[ 4\theta = 2n\pi \pm \frac{\pi}{3} \] Dividing by 4: \[ \theta = \frac{n\pi}{2} \pm \frac{\pi}{12} \] ### Step 2: Solve \( \cos 2\theta = \frac{1}{2} \) Similarly, for \( \cos x = \frac{1}{2} \): \[ x = 2m\pi \pm \frac{\pi}{3} \quad (m \in \mathbb{Z}) \] Substituting \( x = 2\theta \): \[ 2\theta = 2m\pi \pm \frac{\pi}{3} \] Dividing by 2: \[ \theta = m\pi \pm \frac{\pi}{6} \] ### Step 3: Combine the solutions Now we have two sets of solutions: 1. From \( \cos 4\theta = \frac{1}{2} \): \[ \theta = \frac{n\pi}{2} + \frac{\pi}{12} \quad \text{and} \quad \theta = \frac{n\pi}{2} - \frac{\pi}{12} \] 2. From \( \cos 2\theta = \frac{1}{2} \): \[ \theta = m\pi + \frac{\pi}{6} \quad \text{and} \quad \theta = m\pi - \frac{\pi}{6} \] ### Step 4: Identify the solutions from the options Now we will check which of the provided options match the solutions we derived. 1. \( \frac{7\pi}{6} + 2\pi n \) 2. \( \frac{n\pi}{4} \) 3. \( \frac{7\pi}{6} + n\pi \) 4. \( \frac{\pi}{4} + n\pi \) 5. \( 0 + \frac{n\pi}{3} \) 6. \( \frac{11\pi}{6} + m\pi \) 7. \( \frac{3\pi}{4} + n\pi \) 8. \( \frac{11\pi}{6} + 2n\pi \) ### Step 5: Check for specific values - For \( \theta = \frac{n\pi}{2} + \frac{\pi}{12} \) and \( \theta = \frac{n\pi}{2} - \frac{\pi}{12} \), we can find specific values of \( n \) that yield results in the options. - For \( \theta = m\pi + \frac{\pi}{6} \) and \( \theta = m\pi - \frac{\pi}{6} \), we can also find specific values of \( m \). After checking the options, we find that: - \( \frac{7\pi}{6} + n\pi \) can be derived from \( \theta = m\pi + \frac{\pi}{6} \) for specific values of \( m \). - \( \frac{11\pi}{6} + m\pi \) can also be derived from the same. Thus, the solutions that match the derived equations are: - \( \frac{7\pi}{6} + n\pi \) - \( \frac{11\pi}{6} + m\pi \) These are the solutions to the given problem.