Solve the following trigonometric equation assuming that
0 ≤ 𝜃 < 2𝜋.
(Enter your answers as a comma-separated list.)
2 − 3 sin(𝜃) = 1 − 5 sin(𝜃)

We are given the trigonometric equation:

\[
2 - 3\sin(\theta) = 1 - 5\sin(\theta)
\]

### Step 1: Move all terms involving \(\sin(\theta)\) to one side.
Subtract \(1\) from both sides and add \(5\sin(\theta)\) to both sides:

\[
2 - 1 = -5\sin(\theta) + 3\sin(\theta)
\]

This simplifies to:

\[
1 = 2\sin(\theta)
\]

### Step 2: Solve for \(\sin(\theta)\).

Divide both sides by 2:

\[
\sin(\theta) = \frac{1}{2}
\]

### Step 3: Find the values of \(\theta\) that satisfy this equation.

The sine function equals \(\frac{1}{2}\) at the angles:

\[
\theta = \frac{\pi}{6} \quad \text{and} \quad \theta = \frac{5\pi}{6}
\]

### Final Answer:
The solutions to the equation \( 2 - 3\sin(\theta) = 1 - 5\sin(\theta) \) in the interval \( 0 \leq \theta < 2\pi \) are:

\[
\theta = \frac{\pi}{6}, \frac{5\pi}{6}
\]

Would you like more details or have any questions? Here are some related problems to practice:

1. Solve \(3\sin(\theta) - 1 = 0\) for \(0 \leq \theta < 2\pi\).
2. Solve \(2\cos(\theta) = -1\) for \(0 \leq \theta < 2\pi\).
3. Solve \(4\sin(\theta) + 1 = 0\) for \(0 \leq \theta < 2\pi\).
4. Solve \(5\tan(\theta) = \sqrt{3}\) for \(0 \leq \theta < 2\pi\).
5. Solve \(\sin(2\theta) = \frac{1}{2}\) for \(0 \leq \theta < 2\pi\).

**Tip:** When solving trigonometric equations, always consider the periodic nature of the functions and ensure your solutions are within the specified interval.

Learn how to solve the trigonometric equation 2 - 3 sin(theta) = 1 - 5 sin(theta) with a step-by-step solution. This problem involves trigonometric equations and the sine function, suitable for Grades 10-12.

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