sinx+√(3/2)(1–cosx) = 0,

To solve the equation \( \sin x + \sqrt{\frac{3}{2}}(1 - \cos x) = 0 \), let's start by rearranging the equation:

\[ \sin x = -\sqrt{\frac{3}{2}}(1 - \cos x) \]

Now, we can work further with this equation. Recall that \(1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)\). We can replace \(1 - \cos x\) in the original equation:

\[ \sin x = -\sqrt{\frac{3}{2}} \cdot 2 \sin^2\left(\frac{x}{2}\right) \]

This rewrites to:

\[ \sin x = -\sqrt{6} \sin^2\left(\frac{x}{2}\right) \]

Next, we can use the double angle formula for sine, which states that \( \sin x = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \):

\[ 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) = -\sqrt{6} \sin^2\left(\frac{x}{2}\right) \]

Assuming \( \sin\left(\frac{x}{2}\right) \neq 0 \), we can divide both sides by \( \sin\left(\frac{x}{2}\right) \):

\[ 2 \cos\left(\frac{x}{2}\right) = -\sqrt{6} \sin\left(\frac{x}{2}\right) \]

Rearranging gives:

\[ \sqrt{6} \sin\left(\frac{x}{2}\right) + 2 \cos\left(\frac{x}{2}\right) = 0 \]

Now, to find the angles, we can express this in terms of a tangent function:

\[ \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} = -\frac{2}{\sqrt{6}} \implies \tan\left(\frac{x}{2}\right) = -\frac{2}{\sqrt{6}} = -\frac{\sqrt{6}}{3} \]

Now, we need to find the angles corresponding to this tangent value. The tangent function is negative in the second and fourth quadrants. Therefore, we find:

\[ \frac{x}{2} = \tan^{-1}\left(-\frac{\sqrt{6}}{3}\right) + k\pi, \quad k \in \mathbb{Z} \]

Now, we can find specific angles:

\[ \frac{x}{2} = -\theta + k\pi \]

where \( \theta = \tan^{-1}(\frac{\sqrt{6}}{3}) \).

Thus, \(x\) becomes:

\[ x = 2(-\theta + k\pi) = -2\theta + 2k\pi \]

To find numerical approximations, use a calculator to determine \( \theta \):

\[ \theta \approx \tan^{-1}\left(\frac{\sqrt{6}}{3}\right) \approx 0.5236 \text{ radians} \]

Thus, the general solution becomes:

\[ x \approx -2 \cdot 0.5236 + 2k\pi = -1.0472 + 2k\pi \quad \text{ for } k \in \mathbb{Z} \]

This can be expressed as:

\[ x \approx 2k\pi - 1.0472 \]

Thus, the complete solution is:

\[ x = -2\theta + 2k\pi \quad \text{where } k \in \mathbb{Z} \]

This gives all the possible \(x\) values satisfying the original equation.