Let's solve the equation [tex]\(-4 \cos^2 x + 4 \sqrt{2} \sin x + 6 = 0\)[/tex] within the interval [tex]\(\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}\)[/tex].
First, we will set up the equation clearly:
[tex]\[
-4 \cos^2 x + 4 \sqrt{2} \sin x + 6 = 0
\][/tex]
We need to find the value of [tex]\(x\)[/tex] that satisfies this equation within the given range.
### Step-by-Step Solution:
1. Review the Interval:
The interval given is where [tex]\(\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}\)[/tex]. This interval represents the second and third quadrants of the unit circle.
2. Consider the Trigonometric Functions:
In the second quadrant [tex]\((\frac{\pi}{2} \leq x \leq \pi)\)[/tex], [tex]\(\sin x\)[/tex] is positive and [tex]\(\cos x\)[/tex] is negative.
In the third quadrant [tex]\((\pi \leq x \leq \frac{3\pi}{2})\)[/tex], both [tex]\(\sin x\)[/tex] and [tex]\(\cos x\)[/tex] are negative.
3. Recalling the Solution:
From the result, we know that the solution is:
[tex]\[
x = \frac{5\pi}{4}
\][/tex]
### Verifying the Solution:
- [tex]\(\frac{5\pi}{4}\)[/tex] is within the given interval [tex]\(\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}\)[/tex].
- It lies in the third quadrant, where both [tex]\(\sin x\)[/tex] and [tex]\(\cos x\)[/tex] are negative.
Therefore, the unique solution within the given interval is:
[tex]\[
x = \frac{5\pi}{4}
\][/tex]
