Let's solve the equation $2\sin^2\theta - 1 = 0$ within the interval $0 \leq \theta < 2\pi$.

Step 1: Solve the equation for $\sin^2\theta$

$2\sin^2\theta - 1 = 0$

Add 1 to both sides:

$2\sin^2\theta = 1$

Divide by 2:

$\sin^2\theta = \frac{1}{2}$

Step 2: Solve for $\sin\theta$

Take the square root of both sides:

$\sin\theta = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$

Step 3: Find the values of $\theta$ within the interval $0 \leq \theta < 2\pi$

The equation $\sin\theta = \frac{\sqrt{2}}{2}$ has solutions at:

$\theta = \frac{\pi}{4}, \frac{3\pi}{4}$

The equation $\sin\theta = -\frac{\sqrt{2}}{2}$ has solutions at:

$\theta = \frac{5\pi}{4}, \frac{7\pi}{4}$

Final Solution

The solutions to the equation $2\sin^2\theta - 1 = 0$ in the interval $0 \leq \theta < 2\pi$ are:

$\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$

Would you like more details or have any other questions?

Here are 5 related questions:

1. How do you solve trigonometric equations with different intervals?
2. What are the general solutions for $\sin\theta = \frac{\sqrt{2}}{2}$ and $\sin\theta = -\frac{\sqrt{2}}{2}$?
3. Can you explain how to graph the sine function and find its intersections with a given value?
4. How does the unit circle help in finding the solutions to trigonometric equations?
5. What is the significance of each quadrant in the unit circle for trigonometric functions?

Tip: When solving trigonometric equations, always consider the unit circle and the specific interval given to find all possible solutions.