To solve the equation [tex]\(5 \sin x = -3\)[/tex] for [tex]\(0^\circ \leq x \leq 360^\circ\)[/tex], we will follow these steps:
1. Isolate [tex]\(\sin x\)[/tex]:
[tex]\[
\sin x = \frac{-3}{5}
\][/tex]
2. Find the reference angle:
- The value [tex]\(\sin x = \frac{-3}{5}\)[/tex] means that the principal value for [tex]\(x\)[/tex] will be an angle where the sine value is [tex]\(\frac{3}{5}\)[/tex] but in the negative direction.
- Using the inverse sine function ([tex]\(\arcsin\)[/tex]), we can find the principal value:
[tex]\[
\arcsin\left(\frac{3}{5}\right) \approx 36.87^\circ
\][/tex]
- Since the sine function is periodic and symmetrical, and [tex]\(\sin x = \frac{-3}{5}\)[/tex] is negative, the solutions will occur in the 3rd and 4th quadrants.
3. Find all solutions between 0° and 360°:
- In the general context of a unit circle:
- In the 3rd quadrant, [tex]\(x_3 = 180^\circ + 36.87^\circ = 216.87^\circ\)[/tex]
- In the 4th quadrant, [tex]\(x_4 = 360^\circ - 36.87^\circ = 323.13^\circ\)[/tex]
4. Compile the valid solutions within the given range:
- Looking at our principal value and the periodic nature of the sine function,
- [tex]\( x_1 \approx -36.87^\circ \)[/tex] (includes the principal value, but it is not in the valid range of [tex]\(0^\circ \leq x \leq 360^\circ\)[/tex])
- [tex]\( x_2 \approx 216.87^\circ \)[/tex] from 3rd quadrant
- [tex]\( x_3 \approx 143.13^\circ \)[/tex] (Here, [tex]\( 180^\circ - 36.87^\circ = 143.13^\circ \)[/tex] from usual properties of sine, contrary to the initial reference in the angles)
- [tex]\( x_4 \approx 323.13^\circ \)[/tex] from the 4th quadrant, not valid beyond [tex]\(120\degree\)[/tex] in 2nd part.
In conclusion, the solutions to the equation [tex]\(5 \sin x = -3\)[/tex] within the range [tex]\(0^\circ \leq x \leq 360^\circ\)[/tex] are:
[tex]\[
x \approx 216.87^\circ \, \text{and} \, 143.13^\circ
\][/tex]
