Questions: Find the exact solutions of the equation over the given interval. 2 cos^2 theta + 6 = 15/2, 0 ≤ x < 2π

Solution Steps

To solve the given trigonometric equation, we need to isolate the cosine term and then solve for \(\theta\) within the given interval. Here are the steps:

1. Simplify the equation to isolate \(\cos^2 \theta\).
2. Take the square root of both sides to solve for \(\cos \theta\).
3. Determine the values of \(\theta\) that satisfy the equation within the interval \(0 \leq \theta < 2\pi\).

We start with the equation: \[ 2 \cos^2 \theta + 6 = \frac{15}{2} \] Subtracting 6 from both sides gives: \[ 2 \cos^2 \theta = \frac{15}{2} - 6 \] Converting 6 to a fraction: \[ 6 = \frac{12}{2} \] Thus, we have: \[ 2 \cos^2 \theta = \frac{15}{2} - \frac{12}{2} = \frac{3}{2} \] Dividing both sides by 2 results in: \[ \cos^2 \theta = \frac{3}{4} \]

Taking the square root of both sides, we find: \[ \cos \theta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \quad \text{or} \quad \cos \theta = -\sqrt{\frac{3}{4}} = -\frac{\sqrt{3}}{2} \]

We need to find the angles \(\theta\) that satisfy:

1. \(\cos \theta = \frac{\sqrt{3}}{2}\)
2. \(\cos \theta = -\frac{\sqrt{3}}{2}\)

For \(\cos \theta = \frac{\sqrt{3}}{2}\): \[ \theta = \frac{\pi}{6}, \quad \frac{11\pi}{6} \]

For \(\cos \theta = -\frac{\sqrt{3}}{2}\): \[ \theta = \frac{5\pi}{6}, \quad \frac{7\pi}{6} \]

Final Answer