Solve 5Sin O - 3 = 0 for two Values  of 0 degrees to 360 degrees 

To solve the equation:

\[
5\sin \theta - 3 = 0
\]

1. **Isolate** \(\sin \theta\):
   \[
   5\sin \theta = 3
   \]
   \[
   \sin \theta = \frac{3}{5} = 0.6
   \]

2. **Determine values of \(\theta\)** within \(0^\circ\) to \(360^\circ\):

   Since \(\sin \theta = 0.6\), find the reference angle first by using the inverse sine function:

   \[
   \theta = \sin^{-1}(0.6)
   \]
   \[
   \theta \approx 36.87^\circ
   \]

3. **Identify the two angles** in the specified range. Since sine is positive in the first and second quadrants, the solutions are:

   - First quadrant: \(\theta \approx 36.87^\circ\)
   - Second quadrant: \(\theta = 180^\circ - 36.87^\circ \approx 143.13^\circ\)

Thus, the two solutions for \(\theta\) in the range \(0^\circ\) to \(360^\circ\) are approximately:

\[
\theta \approx 36.87^\circ \text{ and } \theta \approx 143.13^\circ
\]

Would you like a detailed breakdown of these steps or have any questions?

---

1. How do we find the general solution for trigonometric equations?
2. What is the method to solve equations involving cosine or tangent functions?
3. How does changing the range affect the solutions for \(\theta\)?
4. How would the solution change if the coefficient of sine was different?
5. What are common applications of solving such trigonometric equations?

**Tip:** When solving trigonometric equations, remember that each function has specific quadrants where it is positive or negative, which affects solution values within a given range.

Solve 5Sin θ - 3 = 0 for two values of θ from 0 degrees to 360 degrees.

Understand how to solve the trigonometric equation 5sin θ - 3 = 0 for values of θ within 0° to 360°. This solution uses trigonometric concepts like the sine function, reference angles, and quadrant analysis to find θ ≈ 36.87° and θ ≈ 143.13°. Ideal for students in Grades 10-12 studying trigonometric equations.

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