Solve 4 − 9sin2(theta) = 0

The given equation is:

$$4 - 9 \sin \theta = 0$$

Subtract 4 from both sides of the equation:

$$-9 \sin \theta = -4$$

Divide both sides by -9:

$$\sin \theta = \frac{-4}{-9}$$ $$\sin \theta = \frac{4}{9}$$

To find $\theta$, we take the inverse sine of $\frac{4}{9}$. Let $\alpha = \arcsin\left(\frac{4}{9}\right)$.
The general solutions for $\sin \theta = k$ are given by:

$$\theta = n \cdot 2\pi + \alpha$$

or

$$\theta = n \cdot 2\pi + (\pi - \alpha)$$

where $n$ is an integer.

Substituting $\alpha = \arcsin\left(\frac{4}{9}\right)$:

$$\theta = n \cdot 2\pi + \arcsin\left(\frac{4}{9}\right)$$

or

$$\theta = n \cdot 2\pi + \pi - \arcsin\left(\frac{4}{9}\right)$$

The solution can be written as:

$$\boxed{\theta = 2n\pi + \arcsin\left(\frac{4}{9}\right) \quad \text{or} \quad \theta = (2n+1)\pi - \arcsin\left(\frac{4}{9}\right), \quad \text{where } n \in \mathbb{Z}}$$