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

We are given the equation: \[ 2\sqrt{2} \cos^2 \theta + (2 - \sqrt{6}) \cos \theta - \sqrt{3} = 0 \] Let \( x = \cos \theta \). Substituting \( x \) into the equation gives: \[ 2\sqrt{2} x^2 + (2 - \sqrt{6}) x - \sqrt{3} = 0 \] This is a quadratic equation in \( x \). We can solve this equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 2\sqrt{2} \), \( b = 2 - \sqrt{6} \), and \( c = -\sqrt{3} \). Substituting these values into the quadratic formula: \[ x = \frac{-(2 - \sqrt{6}) \pm \sqrt{(2 - \sqrt{6})^2 - 4(2\sqrt{2})(-\sqrt{3})}}{2(2\sqrt{2})} \] Simplify this expression and find the roots for \( x \). Since \( \cos \theta \) can only take values between -1 and 1, we will check the valid values of \( x \) that lie within this range. Once we have the valid values for \( x \), we can determine the corresponding values of \( \theta \) in the interval \( [-2\pi, 2\pi] \). Step 1: Find the roots for \( x \). After simplifying the quadratic equation, we find the roots for \( x \). Step 2: Find the values of \( \theta \). Each valid value of \( x = \cos \theta \) corresponds to two values of \( \theta \) in the interval \( [-2\pi, 2\pi] \). Thus, the total number of solutions is 4.