Solving equations Solve each equation.
√2 sin 3Θ + 1 = 2, 0 ≤ Θ ≤ π

Verified step by step guidance

1. Step 1: Start by isolating the trigonometric function. Subtract 1 from both sides of the equation to get \( \sqrt{2} \sin 3\Theta = 1 \).
2. Step 2: Divide both sides by \( \sqrt{2} \) to solve for \( \sin 3\Theta \). This gives \( \sin 3\Theta = \frac{1}{\sqrt{2}} \).
3. Step 3: Recognize that \( \sin 3\Theta = \frac{1}{\sqrt{2}} \) corresponds to the standard angle \( \frac{\pi}{4} \) in the unit circle, where sine is positive.
4. Step 4: Solve for \( 3\Theta \) by setting \( 3\Theta = \frac{\pi}{4} + 2k\pi \) and \( 3\Theta = \pi - \frac{\pi}{4} + 2k\pi \) for integer \( k \), since sine is positive in the first and second quadrants.
5. Step 5: Divide each solution by 3 to solve for \( \Theta \), ensuring that the solutions fall within the given interval \( 0 \leq \Theta \leq \pi \).

Key Concepts

Trigonometric Functions

Trigonometric functions, such as sine and cosine, relate angles to ratios of sides in right triangles. In this context, the sine function, denoted as sin(Θ), is crucial for solving the equation involving the angle Θ. Understanding the properties and values of these functions is essential for finding solutions within the specified interval.

Inverse Trigonometric Functions

Inverse trigonometric functions, like arcsin, allow us to determine the angle corresponding to a given sine value. When solving equations involving trigonometric functions, we often need to apply these inverses to find the angle Θ that satisfies the equation. This concept is vital for extracting solutions from the results of the trigonometric calculations.

Interval Notation

Interval notation specifies the range of values for which a solution is valid. In this problem, the interval 0 ≤ Θ ≤ π indicates that we are only interested in solutions for Θ within this range. Understanding how to interpret and apply interval notation is important for ensuring that the solutions found are relevant to the problem's constraints.