We are asked to find the number of solutions for the equation $2\sqrt{2}\cos^2\theta+(2-\sqrt{6}) \cos\theta-\sqrt{3}=0$ in the interval $\theta \in [-2\pi, 2\pi]$.

Solving the Trigonometric Equation

Step 1: Substitute and form a quadratic equation

Let $x = \cos\theta$. The equation transforms into a quadratic equation in terms of $x$:

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

Step 2: Solve the quadratic equation for $x$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2\sqrt{2}$, $b = 2-\sqrt{6}$, and $c = -\sqrt{3}$:

$ x = \frac{-(2-\sqrt{6}) \pm \sqrt{(2-\sqrt{6})^2 - 4(2\sqrt{2})(-\sqrt{3})}}{2(2\sqrt{2})} $

First, calculate the discriminant $\Delta = b^2 - 4ac$:

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

Simplify the square root of the discriminant:

$ \sqrt{\Delta} = \sqrt{10 + 4\sqrt{6}} = \sqrt{10 + 2\sqrt{24}} $

We look for two numbers that multiply to 24 and add to 10, which are 6 and 4. So,

$ \sqrt{10 + 2\sqrt{24}} = \sqrt{6} + \sqrt{4} = \sqrt{6} + 2 $

Now substitute back into the quadratic formula:

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

This gives two possible values for $x$:

• $x_1 = \frac{-2+\sqrt{6} + (\sqrt{6} + 2)}{4\sqrt{2}} = \frac{2\sqrt{6}}{4\sqrt{2}} = \frac{\sqrt{3}}{2}$
• $x_2 = \frac{-2+\sqrt{6} - (\sqrt{6} + 2)}{4\sqrt{2}} = \frac{-4}{4\sqrt{2}} = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$

Step 3: Find the values of $\theta$ in the interval $[-2\pi, 2\pi]$

We need to solve $\cos\theta = \frac{\sqrt{3}}{2}$ and $\cos\theta = -\frac{\sqrt{2}}{2}$ for $\theta \in [-2\pi, 2\pi]$.

Case 1: $\cos\theta = \frac{\sqrt{3}}{2}$

The general solution is $\theta = 2n\pi \pm \frac{\pi}{6}$, where $n$ is an integer.

For $n=0$, $\theta = \pm \frac{\pi}{6}$. Both are in $[-2\pi, 2\pi]$.

For $n=1$, $\theta = 2\pi \pm \frac{\pi}{6}$. This gives $\theta = \frac{11\pi}{6}$ (in $[0, 2\pi]$) and $\theta = \frac{13\pi}{6}$ (outside the interval).

For $n=-1$, $\theta = -2\pi \pm \frac{\pi}{6}$. This gives $\theta = -\frac{11\pi}{6}$ (in $[-2\pi, 0]$) and $\theta = -\frac{13\pi}{6}$ (outside the interval).

Solutions in $[-2\pi, 2\pi]$ are: $\frac{\pi}{6}, \frac{11\pi}{6}, -\frac{\pi}{6}, -\frac{11\pi}{6}$. (4 solutions)

Case 2: $\cos\theta = -\frac{\sqrt{2}}{2}$

The general solution is $\theta = 2n\pi \pm \frac{3\pi}{4}$, where $n$ is an integer.

For $n=0$, $\theta = \pm \frac{3\pi}{4}$. Both are in $[-2\pi, 2\pi]$.

For $n=1$, $\theta = 2\pi \pm \frac{3\pi}{4}$. This gives $\theta = \frac{5\pi}{4}$ (in $[0, 2\pi]$) and $\theta = \frac{11\pi}{4}$ (outside the interval).

For $n=-1$, $\theta = -2\pi \pm \frac{3\pi}{4}$. This gives $\theta = -\frac{5\pi}{4}$ (in $[-2\pi, 0]$) and $\theta = -\frac{11\pi}{4}$ (outside the interval).

Solutions in $[-2\pi, 2\pi]$ are: $\frac{3\pi}{4}, \frac{5\pi}{4}, -\frac{3\pi}{4}, -\frac{5\pi}{4}$. (4 solutions)

Step 4: Count the total number of distinct solutions

Combining the solutions from both cases, we have the set:

$ \{ -\frac{11\pi}{6}, -\frac{5\pi}{4}, -\frac{3\pi}{4}, -\frac{\pi}{6}, \frac{\pi}{6}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{11\pi}{6} \} $

All 8 solutions are distinct and lie within the specified interval $[-2\pi, 2\pi]$.

Therefore, the total number of solutions is $4 + 4 = 8$.