QUESTION \( 1(10 \) marks) Solve \( 3 \cos 2 \theta+0.6=0 \) for \( 0 \leq \theta<2 \pi \)

To solve the equation \( 3 \cos 2 \theta + 0.6 = 0 \), we first isolate \( \cos 2 \theta \): \[ 3 \cos 2 \theta = -0.6 \] \[ \cos 2 \theta = -0.2 \] Next, we take the inverse cosine to find \( 2\theta \): \[ 2\theta = \cos^{-1}(-0.2) \] Calculating the angles gives us: \[ 2\theta \approx 1.7628 \quad \text{and} \quad 2\theta \approx 4.4402 \quad \text{(since cosine is negative in the second and third quadrants)} \] Dividing by 2 to find \( \theta \): 1. First angle: \[ \theta \approx \frac{1.7628}{2} \approx 0.8814 \] 2. Second angle: \[ \theta \approx \frac{4.4402}{2} \approx 2.2201 \] Next, we consider that the cosine function is periodic with a period of \( 2\pi \). Thus, we can add \( \pi \) to the angle since \( 2\theta \) will have a cycle of \( 2\pi \): For \( 2\theta = 1.7628 + 2k\pi \): \[ 2\theta = 1.7628 + \pi \Rightarrow 2\theta \approx 1.7628 + 3.1416 \approx 4.9044 \Rightarrow \theta \approx 2.4522 \quad (k=0) \] For \( 2\theta = 4.4402 + 2k\pi \): \[ 2\theta = 4.4402 + \pi \Rightarrow 2\theta \approx 4.4402 + 3.1416 \approx 7.5818 \Rightarrow \theta \approx 3.7909 \quad (k=0) \] Collecting all possible solutions for \( \theta \) within the interval \( [0, 2\pi) \): \[ \theta \approx 0.8814, \quad 2.2201, \quad 2.4522, \quad 3.7909 \] Hence, the final solutions are: \[ \theta \approx 0.8814, \quad 2.2201, \quad 2.4522, \quad 3.7909 \]