Solving equations Solve each equation.


sin² 2Θ = 1/2, -π/2 ≤ Θ ≤ π/2

Verified step by step guidance

1. Step 1: Recognize that the equation is in the form of \( \sin^2(2\Theta) = \frac{1}{2} \). To solve for \( \Theta \), first take the square root of both sides to get \( \sin(2\Theta) = \pm \frac{1}{\sqrt{2}} \).
2. Step 2: Recall that \( \sin(2\Theta) = \pm \frac{1}{\sqrt{2}} \) corresponds to angles where the sine function equals \( \frac{1}{\sqrt{2}} \) or \( -\frac{1}{\sqrt{2}} \). These angles are \( \frac{\pi}{4}, \frac{3\pi}{4}, -\frac{\pi}{4}, \) and \( -\frac{3\pi}{4} \).
3. Step 3: Set \( 2\Theta = \frac{\pi}{4}, \frac{3\pi}{4}, -\frac{\pi}{4}, \) and \( -\frac{3\pi}{4} \) to find the possible values of \( 2\Theta \).
4. Step 4: Solve for \( \Theta \) by dividing each angle by 2. This gives \( \Theta = \frac{\pi}{8}, \frac{3\pi}{8}, -\frac{\pi}{8}, \) and \( -\frac{3\pi}{8} \).
5. Step 5: Verify that each solution for \( \Theta \) falls within the given interval \( -\frac{\pi}{2} \leq \Theta \leq \frac{\pi}{2} \).