Question

Number of solutions in the interval \( [-2\pi, 2\pi] \) for the equation: \[ 2\sqrt{2} \cos^2 \theta + (2 - \sqrt{6}) \cos \theta - \sqrt{3} = 0 \]

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

For quadratic equations involving trigonometric functions, always ensure the solutions lie within the range for the trigonometric function (e.g., \( -1 \leq \cos \theta \leq 1 \)).

Answer 1

The Correct Option is C

The given equation is: \[ 2\sqrt{2} \cos^2 \theta + (2 - \sqrt{6}) \cos \theta - \sqrt{3} = 0 \] Let \( x = \cos \theta \). Substituting \( x \) yields: \[ 2\sqrt{2} x^2 + (2 - \sqrt{6}) x - \sqrt{3} = 0 \] This is a quadratic equation in \( x \). Applying the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 2\sqrt{2} \), \( b = 2 - \sqrt{6} \), and \( c = -\sqrt{3} \): \[ x = \frac{-(2 - \sqrt{6}) \pm \sqrt{(2 - \sqrt{6})^2 - 4(2\sqrt{2})(-\sqrt{3})}}{2(2\sqrt{2})} \] Simplify to find the roots for \( x \). Since \( \cos \theta \) is restricted to the interval [-1, 1], we will select valid \( x \) values within this range. For each valid \( x \), determine the corresponding \( \theta \) values in the interval \( [-2\pi, 2\pi] \). Step 1: Solve the quadratic equation for \( x \). Step 2: For each valid \( x \in [-1, 1] \), find the values of \( \theta \) in \( [-2\pi, 2\pi] \) such that \( \cos \theta = x \). Each valid \( x \) yields two \( \theta \) values in the specified interval. The total number of solutions is 4.