To simplify the given trigonometric equation, we can make a substitution. Let x = cosθ. This transforms the equation into a quadratic equation in terms of x.

We use the quadratic formula x = (-b ± √(b² - 4ac))/(2a) to find the values of x. Here, a = 2√2, b = 2-√6, and c = -√3.

First, we calculate the discriminant b² - 4ac. Then, we simplify the square root term. Notice that 10 + 4√6 can be written as (√4 + √6)². Finally, we find the two possible values for x.

For cosθ = (√3)/(2), the principal value is (π)/(6). Since the cosine function is positive in the first and fourth quadrants, and the interval is [-2π, 2π], the solutions are (π)/(6), -(π)/(6), (11π)/(6) (which is 2π - (π)/(6)), and -(11π)/(6) (which is -2π + (π)/(6)). This gives 4 solutions.

For cosθ = (√3)/(2), the principal value is (π)/(6). Since the cosine function is positive in the first and fourth quadrants, and the interval is [-2π, 2π], the solutions are (π)/(6), -(π)/(6), (11π)/(6) (which is 2π - (π)/(6)), and -(11π)/(6) (which is -2π + (π)/(6)). This gives 4 solutions.

For cosθ = -(1)/(√2), the principal value is (3π)/(4). Since the cosine function is negative in the second and third quadrants, and the interval is [-2π, 2π], the solutions are (3π)/(4), -(3π)/(4), (5π)/(4) (which is 2π - (3π)/(4) is incorrect, it should be 2π - (3π)/(4) is not in the range, it should be π + (π)/(4)), and -(5π)/(4). This gives 4 solutions.

For cosθ = -(1)/(√2), the principal value is (3π)/(4). Since the cosine function is negative in the second and third quadrants, and the interval is [-2π, 2π], the solutions are (3π)/(4), -(3π)/(4), (5π)/(4) (which is 2π - (3π)/(4) is incorrect, it should be 2π - (3π)/(4) is not in the range, it should be π + (π)/(4)), and -(5π)/(4). This gives 4 solutions.

Combining the solutions from both cases, cosθ = (√3)/(2) and cosθ = -(1)/(√2), we get a total of 4 + 4 = 8 distinct solutions in the interval [-2π, 2π].

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 equal to:

Answer: 8

Step-by-step solution

Step 1: Substitute and Simplify

To simplify the given trigonometric equation, we can make a substitution. Let $x = \cos\theta$ . This transforms the equation into a quadratic equation in terms of $x$ .

Step 2: Solve the Quadratic Equation

We use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the values of $x$ . Here, $a = 2\sqrt{2}$ , $b = 2-\sqrt{6}$ , and $c = -\sqrt{3}$ .

Step 3: Calculate Discriminant and Roots

First, we calculate the discriminant $b^2 - 4ac$ . Then, we simplify the square root term. Notice that $10 + 4\sqrt{6}$ can be written as $(\sqrt{4} + \sqrt{6})^2$ . Finally, we find the two possible values for $x$ .

Step 4: Find Solutions for $\cos\theta = \frac{\sqrt{3}}{2}$

For $\cos\theta = \frac{\sqrt{3}}{2}$ , the principal value is $\frac{\pi}{6}$ . Since the cosine function is positive in the first and fourth quadrants, and the interval is $[-2\pi, 2\pi]$ , the solutions are $\frac{\pi}{6}$ , $-\frac{\pi}{6}$ , $\frac{11\pi}{6}$ (which is $2\pi - \frac{\pi}{6}$ ), and $-\frac{11\pi}{6}$ (which is $-2\pi + \frac{\pi}{6}$ ). This gives 4 solutions.

Step 5: Find Solutions for $\cos\theta = -\frac{1}{\sqrt{2}}$

For $\cos\theta = -\frac{1}{\sqrt{2}}$ , the principal value is $\frac{3\pi}{4}$ . Since the cosine function is negative in the second and third quadrants, and the interval is $[-2\pi, 2\pi]$ , the solutions are $\frac{3\pi}{4}$ , $-\frac{3\pi}{4}$ , $\frac{5\pi}{4}$ (which is $2\pi - \frac{3\pi}{4}$ is incorrect, it should be $2\pi - \frac{3\pi}{4}$ is not in the range, it should be $\pi + \frac{\pi}{4}$ ), and $-\frac{5\pi}{4}$ . This gives 4 solutions.

Step 6: Total Number of Solutions

Combining the solutions from both cases, $\cos\theta = \frac{\sqrt{3}}{2}$ and $\cos\theta = -\frac{1}{\sqrt{2}}$ , we get a total of $4 + 4 = 8$ distinct solutions in the interval $[-2\pi, 2\pi]$ .

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If θ∈[−2π,2π]\theta \in [-2\pi, 2\pi]θ∈[−2π,2π], then the number of solutions of 22cos⁡2θ+(2−6)cos⁡θ−3=02\sqrt{2}\cos^2\theta + (2-\sqrt{6})\cos\theta - \sqrt{3} = 022​cos2θ+(2−6​)cosθ−3​=0 is equal to: