We are asked to solve the equation:

$2\cdot \sin(x) + \sqrt{3} = 0$

Step 1: Isolate $\sin(x)$

First, we isolate $\sin(x)$ by subtracting $\sqrt{3}$ from both sides:

$2\cdot \sin(x) = -\sqrt{3}$

Now, divide both sides by 2:

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

Step 2: Determine the angle(s)

Next, we need to find the values of $x$ for which $\sin(x) = -\frac{\sqrt{3}}{2}$.

The reference angle where $\sin(\theta) = \frac{\sqrt{3}}{2}$ is $\theta = \frac{\pi}{3}$. Since we are looking for $\sin(x) = -\frac{\sqrt{3}}{2}$, we need to consider angles where the sine function is negative. These occur in the third and fourth quadrants.

Thus, the angles satisfying $\sin(x) = -\frac{\sqrt{3}}{2}$ are:

• In the third quadrant: $x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$
• In the fourth quadrant: $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$

Step 3: General solution

Since the sine function is periodic with a period of $2\pi$, the general solutions are:

$x = \frac{4\pi}{3} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{3} + 2n\pi$

where $n$ is any integer.

Final Answer:

The general solutions to the equation $2\cdot \sin(x) + \sqrt{3} = 0$ are:

$x = \frac{4\pi}{3} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{3} + 2n\pi, \quad n \in \mathbb{Z}.$

Would you like more details or have any questions?

Further questions:

1. How does the periodicity of the sine function affect the general solution?
2. What other trigonometric functions might help solve similar equations?
3. How would the solutions change if the equation involved $\cos(x)$ instead of $\sin(x)$?
4. Can we use a calculator to approximate the values of $x$?
5. How does the unit circle help in finding solutions for trigonometric equations?

Tip: When solving trigonometric equations, always look for the reference angle first and then determine the correct quadrant based on the sign of the function.