The problem consists of two questions. Question 9: Determine the quadrants in which the solutions to the equation $2\sin x + \sqrt{3} = 0$ lie. Question 10: If $\cos x = -\frac{1}{\sqrt{2}}$ and $0 < x < 2\pi$, find a possible value of $x$.

1. Problem Description

The problem consists of two questions.

Question 9: Determine the quadrants in which the solutions to the equation $2\sin x + \sqrt{3} = 0$ lie.

Question 10: If $\cos x = -\frac{1}{\sqrt{2}}$ and $0 < x < 2\pi$, find a possible value of $x$.

2. Solution Steps

Question 9:

First, solve the equation $2\sin x + \sqrt{3} = 0$ for $\sin x$:

$2\sin x = -\sqrt{3}$

$\sin x = -\frac{\sqrt{3}}{2}$

Since $\sin x$ is negative, the angle $x$ must lie in either quadrant III or quadrant IV.

Question 10:

We are given that $\cos x = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$ and $0 < x < 2\pi$. Since the cosine is negative, $x$ must lie in either quadrant II or quadrant III. We know that $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$. Therefore, the reference angle is $\frac{\pi}{4}$.

In quadrant II, $x = \pi - \frac{\pi}{4} = \frac{4\pi}{4} - \frac{\pi}{4} = \frac{3\pi}{4}$.

In quadrant III, $x = \pi + \frac{\pi}{4} = \frac{4\pi}{4} + \frac{\pi}{4} = \frac{5\pi}{4}$.

Since $\frac{3\pi}{4}$ is among the possible answers, we choose that.

3. Final Answer

Question 9: III, IV

Question 10: $\frac{3\pi}{4}$