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

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Feb 03,2025

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The solutions for \( \theta \) are approximately \( 0.886 \), \( 2.256 \), \( 4.027 \), and \( 5.397 \) radians.

Solución

To solve the equation \( 3 \cos(2\theta) + 0.6 = 0 \) for \( 0 \leq \theta < 2\pi \), follow these steps: ### Step 1: Isolate the Trigonometric Function Start by isolating the cosine term: \[ 3 \cos(2\theta) + 0.6 = 0 \] Subtract 0.6 from both sides: \[ 3 \cos(2\theta) = -0.6 \] Divide both sides by 3: \[ \cos(2\theta) = -0.2 \] ### Step 2: Find the General Solutions for \( 2\theta \) The equation \( \cos(2\theta) = -0.2 \) can be solved by finding the general solutions for \( 2\theta \): \[ 2\theta = \cos^{-1}(-0.2) + 2k\pi \quad \text{and} \quad 2\theta = -\cos^{-1}(-0.2) + 2k\pi \quad \text{for integer } k \] Calculate \( \cos^{-1}(-0.2) \): \[ \cos^{-1}(-0.2) \approx 1.772 \text{ radians} \] Thus, the general solutions for \( 2\theta \) are: \[ 2\theta \approx 1.772 + 2k\pi \quad \text{and} \quad 2\theta \approx -1.772 + 2k\pi \] ### Step 3: Solve for \( \theta \) Within the Specified Interval Divide both sides by 2 to solve for \( \theta \): \[ \theta \approx \frac{1.772}{2} + k\pi \quad \text{and} \quad \theta \approx \frac{-1.772}{2} + k\pi \] Simplify: \[ \theta \approx 0.886 + k\pi \quad \text{and} \quad \theta \approx -0.886 + k\pi \] Now, find all solutions within \( 0 \leq \theta < 2\pi \): 1. For \( k = 0 \): - \( \theta \approx 0.886 \) radians - \( \theta \approx -0.886 + \pi \approx 2.255 \) radians 2. For \( k = 1 \): - \( \theta \approx 0.886 + \pi \approx 4.027 \) radians - \( \theta \approx -0.886 + 2\pi \approx 5.397 \) radians ### Step 4: List All Solutions The solutions within the interval \( 0 \leq \theta < 2\pi \) are: \[ \theta \approx 0.886 \text{ radians},\ 2.256 \text{ radians},\ 4.027 \text{ radians},\ 5.397 \text{ radians} \] ### Final Answer The solutions for \( \theta \) in the interval \( 0 \leq \theta < 2\pi \) are approximately: \[ \theta \approx 0.886,\ 2.256,\ 4.027,\ \text{and } 5.397 \text{ radians} \]

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To solve the equation \( 3 \cos 2\theta + 0.6 = 0 \), we first isolate the cosine term: \[ 3 \cos 2\theta = -0.6 \] \[ \cos 2\theta = -0.2 \] Next, we need to find the values of \( 2\theta \) that satisfy this equation. The cosine function is negative in the second and third quadrants. Thus, we can use the inverse cosine function: \[ 2\theta = \cos^{-1}(-0.2) \quad \text{or} \quad 2\theta = 2\pi - \cos^{-1}(-0.2) \] Calculating \( \cos^{-1}(-0.2) \): \[ \cos^{-1}(-0.2) \approx 1. مثəl859 \] So our first solutions for \( 2\theta \): \[ 2\theta \approx 1.823 \quad \text{(2nd quadrant)} \] And for the 3rd quadrant: \[ 2\theta \approx 2\pi - 1.823 \approx 4.460 \] Now, since \( 2\theta \) can be expressed periodically, we add \( 2\pi k \) for \( k \in \mathbb{Z} \), giving us: 1. \( 2\theta = 1.823 + 2\pi k \) 2. \( 2\theta = 4.460 + 2\pi k \) Dividing by 2, we get: 1. \( \theta \approx 0.9115 + \pi k \) 2. \( \theta \approx 2.2300 + \pi k \) Now, we substitute \( k = 0 \) and \( k = 1 \) to find the values of \( \theta \) within the interval \( [0, 2\pi) \): For \( k = 0 \): - \( \theta \approx 0.9115 \) - \( \theta \approx 2.2300 \) For \( k = 1 \): - \( \theta \approx 0.9115 + \pi \approx 4.0520 \) - \( \theta \approx 2.2300 + \pi \approx 5.3716 \) Finally, the solutions for \( \theta \) in \( [0, 2\pi) \) are approximately: \[ \theta \approx 0.9115, \quad 2.2300, \quad 4.0520, \quad 5.3716 \]

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