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

We are asked to find the total number 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:

The given equation is a quadratic equation in the form of \( ax^2 + bx + c = 0 \), where \( x = \cos\theta \). We will solve this quadratic equation for \( \cos\theta \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

After finding the possible values for \( \cos\theta \), we will determine the corresponding values of \( \theta \) in the interval \( [-2\pi, 2\pi] \). 
The number of solutions will be the count of all such distinct values of \( \theta \).

Step-by-Step Solution:

Let \( x = \cos\theta \). The given equation can be rewritten as a quadratic equation in \( x \):

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

Here, the coefficients are \( a = 2\sqrt{2} \), \( b = 2 - \sqrt{6} \), and \( c = -\sqrt{3} \). 
We apply the quadratic formula to solve for \( x \).

First, we calculate the discriminant, \( \Delta = b^2 - 4ac \):

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

To simplify \( \sqrt{\Delta} \), we try to express \( 10 + 4\sqrt{6} \) as a perfect square of the form \( (p+q)^2 \). We can write \( 10 + 4\sqrt{6} = 10 + 2\sqrt{24} \). 
We look for two numbers whose sum is 10 and product is 24. 
These numbers are 6 and 4. So,

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

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

Now, we substitute this back into the quadratic formula:

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

This gives two possible values for \( x = \cos\theta \):

Case 1: Using the '+' sign.

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

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 we find the number of solutions for each case in the interval \( [-2\pi, 2\pi] \).

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

In the interval \( [0, 2\pi] \), \( \cos\theta \) is positive in the first and fourth quadrants. 
The solutions are \( \theta = \frac{\pi}{6} \) and \( \theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \). 
In the interval \( [-2\pi, 0] \), the solutions are \( \theta = -\frac{\pi}{6} \) and \( \theta = -2\pi + \frac{\pi}{6} = -\frac{11\pi}{6} \). 
Thus, there are 4 solutions for this case: \( \frac{\pi}{6}, \frac{11\pi}{6}, -\frac{\pi}{6}, -\frac{11\pi}{6} \).

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

In the interval \( [0, 2\pi] \), \( \cos\theta \) is negative in the second and third quadrants. 
The solutions are \( \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \) and \( \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \). 
In the interval \( [-2\pi, 0] \), the solutions are \( \theta = -\frac{3\pi}{4} \) and \( \theta = -\frac{5\pi}{4} \). 
Thus, there are 4 solutions for this case: \( \frac{3\pi}{4}, \frac{5\pi}{4}, -\frac{3\pi}{4}, -\frac{5\pi}{4} \).

Final Computation & Result:

The total number of solutions is the sum of the number of solutions from both cases. 
Since the values \( \frac{\sqrt{3}}{2} \) and \( -\frac{1}{\sqrt{2}} \) are distinct, there are no overlapping solutions.

Number of solutions from Case 1 = 4.

Number of solutions from Case 2 = 4.

Total number of solutions = 4 + 4 = 8.

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

Answer 2

Let \( x = \cos\theta \). Then the equation becomes: \[ 2\sqrt{2}x^2 + (2 - \sqrt{6})x - \sqrt{3} = 0 \] Solve the quadratic: \[ (2x - \sqrt{3})(\sqrt{2}x + 1) = 0 \Rightarrow x = \frac{\sqrt{3}}{2},\ -\frac{1}{\sqrt{2}} \] Each valid cosine value gives 4 solutions in \( [-2\pi,\ 2\pi] \), so total = 8 solutions.