Solve the equation 3 an2 heta − 1 = 0 for heta in the range 0^ <= theta <= 90^.

The given equation is:

$$3\tan^2\theta - 1 = 0$$

Add 1 to both sides:

$$3\tan^2\theta = 1$$

Divide by 3:

$$\tan^2\theta = \frac{1}{3}$$

$$\tan\theta = \pm\sqrt{\frac{1}{3}}$$ $$\tan\theta = \pm\frac{1}{\sqrt{3}}$$

The problem specifies that $0^\circ \leq \theta \leq 90^\circ$. In this range (the first quadrant), the tangent function is positive. Therefore, we take the positive value:

$$\tan\theta = \frac{1}{\sqrt{3}}$$

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$.
So,

$$\theta = 30^\circ$$

This value is within the specified range $0^\circ \leq \theta \leq 90^\circ$.

The solution is $30^\circ$.