To solve the given problem, we start with the equation: \(\cos^2\theta - \sin^2\theta - 3\cos\theta + 2 = 0\). We will use the trigonometric identity \(\cos^2\theta - \sin^2\theta = \cos(2\theta)\) to simplify the equation.

1. Applying the identity, we have: \(\cos(2\theta) - 3\cos\theta + 2 = 0\).
2. Let \(x = \cos\theta\). Then, \(\cos(2\theta) = 2x^2 - 1\). So, the equation becomes:

\(2x^2 - 1 - 3x + 2 = 0\)

1. Simplify the equation: \(2x^2 - 3x + 1 = 0\).
2. This is a quadratic equation in \(x\). Solving it using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -3\), and \(c = 1\).

The roots are given by:

\(x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2}\)

\(x = \frac{3 \pm \sqrt{9 - 8}}{4}\)

\(x = \frac{3 \pm 1}{4}\)

Therefore, \(\cos\theta\) can be:

• \(x = \frac{4}{4} = 1\)
• \(x = \frac{2}{4} = \frac{1}{2}\)

Since \(0^\circ < \theta < 90^\circ\), we can eliminate \(\cos\theta = 1\) because this would correspond to \(\theta = 0^\circ\). Thus, \(\cos\theta = \frac{1}{2}\), corresponding to \(\theta = 60^\circ\).

1. Now, we want to find the value of \(5\cos\theta - \frac{\tan\theta}{2}\).

First, calculate each term:

1. \(\cos\theta = \frac{1}{2}\), hence \(5\cos\theta = 5 \times \frac{1}{2} = \frac{5}{2}\).
2. Since \(\theta = 60^\circ\), \(\tan\theta = \sqrt{3}\). Therefore,

\(\frac{\tan\theta}{2} = \frac{\sqrt{3}}{2}\)

1. Substituting these values back, we get:

\(5\cos\theta - \frac{\tan\theta}{2} = \frac{5}{2} - \frac{\sqrt{3}}{2} = \frac{5 - \sqrt{3}}{2}\)

Thus, the value of \(5\cos\theta - \frac{\tan\theta}{2}\) is \(\frac{5-\sqrt{3}}{2}\), which matches the correct answer.