Question

If $ \theta \in [-2\pi,\ 2\pi] $, then the number of solutions of $$ 2\sqrt{2} \cos^2\theta + (2 - \sqrt{6}) \cos\theta - \sqrt{3} = 0 $$ is:

• A. 12
• B. 6
• C. 8
• D. 10

Hint:

Transform trig equations into algebraic form using identities to find roots easily.

Answer 1

The Correct Option is C

The objective is to determine the total count of solutions for the trigonometric equation \( 2\sqrt{2} \cos^2\theta + (2 - \sqrt{6}) \cos\theta - \sqrt{3} = 0 \) within the interval \( \theta \in [-2\pi, 2\pi] \).

Concept Used:

This problem involves solving a quadratic equation in terms of \( \cos\theta \). We will use the standard quadratic formula to find the values of \( \cos\theta \) and then determine the corresponding \( \theta \) values within the specified interval.

The quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Following the calculation of \( \cos\theta \) values, we will identify all distinct \( \theta \) values within \( [-2\pi, 2\pi] \).

Step-by-Step Solution:

Let \( x = \cos\theta \). The equation transforms into a quadratic form:

\[ 2\sqrt{2} x^2 + (2 - \sqrt{6}) x - \sqrt{3} = 0 \]

The coefficients are \( a = 2\sqrt{2} \), \( b = 2 - \sqrt{6} \), and \( c = -\sqrt{3} \).

Calculate the discriminant \( \Delta = b^2 - 4ac \):

\[ \Delta = (2 - \sqrt{6})^2 - 4(2\sqrt{2})(-\sqrt{3}) \] \[ \Delta = (4 - 4\sqrt{6} + 6) + 8\sqrt{6} \] \[ \Delta = 10 + 4\sqrt{6} \]

Simplify \( \sqrt{\Delta} \). Express \( 10 + 4\sqrt{6} \) as \( 10 + 2\sqrt{24} \). Find two numbers that sum to 10 and multiply to 24; these are 6 and 4.

\[ 10 + 4\sqrt{6} = 6 + 4 + 2\sqrt{6 \cdot 4} = (\sqrt{6})^2 + (2)^2 + 2(2)\sqrt{6} = (2 + \sqrt{6})^2 \]

Thus, \( \sqrt{\Delta} = 2 + \sqrt{6} \).

Apply the quadratic formula:

\[ x = \frac{-(2 - \sqrt{6}) \pm (2 + \sqrt{6})}{2(2\sqrt{2})} = \frac{\sqrt{6} - 2 \pm (2 + \sqrt{6})}{4\sqrt{2}} \]

Two possible values for \( x = \cos\theta \) emerge:

Case 1: Using the '+' sign.

\[ \cos\theta = \frac{(\sqrt{6} - 2) + (2 + \sqrt{6})}{4\sqrt{2}} = \frac{2\sqrt{6}}{4\sqrt{2}} = \frac{\sqrt{3}}{2} \]

Case 2: Using the '-' sign.

\[ \cos\theta = \frac{(\sqrt{6} - 2) - (2 + \sqrt{6})}{4\sqrt{2}} = \frac{-4}{4\sqrt{2}} = -\frac{1}{\sqrt{2}} \]

Now, determine the solutions for \( \theta \) in \( [-2\pi, 2\pi] \) for each case.

For \( \cos\theta = \frac{\sqrt{3}}{2} \):

In \( [0, 2\pi] \), solutions are \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{11\pi}{6} \).

In \( [-2\pi, 0] \), solutions are \( \theta = -\frac{\pi}{6} \) and \( \theta = -\frac{11\pi}{6} \).

Total of 4 solutions: \( \frac{\pi}{6}, \frac{11\pi}{6}, -\frac{\pi}{6}, -\frac{11\pi}{6} \).

For \( \cos\theta = -\frac{1}{\sqrt{2}} \):

In \( [0, 2\pi] \), solutions are \( \theta = \frac{3\pi}{4} \) and \( \theta = \frac{5\pi}{4} \).

In \( [-2\pi, 0] \), solutions are \( \theta = -\frac{3\pi}{4} \) and \( \theta = -\frac{5\pi}{4} \).

Total of 4 solutions: \( \frac{3\pi}{4}, \frac{5\pi}{4}, -\frac{3\pi}{4}, -\frac{5\pi}{4} \).

Final Computation & Result:

Sum the solutions from both cases. Since \( \frac{\sqrt{3}}{2} \) and \( -\frac{1}{\sqrt{2}} \) are distinct, the solutions are unique.

Solutions from Case 1: 4.

Solutions from Case 2: 4.

Total solutions = 4 + 4 = 8.

The total number of solutions for the given equation in the interval \( [-2\pi, 2\pi] \) is 8.