In the interval [tex]\(0^{\circ} \ \textless \ x \ \textless \ 360^{\circ}\)[/tex], find the values of [tex]\(x\)[/tex] for which [tex]\(\sin x = 0.6691\)[/tex]. Give your answers to the nearest degree. [tex]\(\square\)[/tex]



Answer :

To find the values of [tex]\( x \)[/tex] in the interval [tex]\( 0^{\circ} < x < 360^{\circ} \)[/tex] for which [tex]\( \sin x = 0.6691 \)[/tex], we follow these steps:

1. Determine the principal value:
- Start by finding the principal value for [tex]\( x \)[/tex] when [tex]\( \sin x = 0.6691 \)[/tex]. This involves using the arcsine function, [tex]\( \sin^{-1} \)[/tex], to find an angle whose sine is 0.6691.
- The value is approximately [tex]\( 42^\circ \)[/tex].

2. Identify the second solution in one full circle:
- Since the sine function is positive in both the first and second quadrants, there is another angle in the second quadrant where [tex]\( \sin x = 0.6691 \)[/tex].
- This angle can be found by subtracting the principal angle from [tex]\( 180^\circ \)[/tex], i.e., [tex]\( 180^\circ - 42^\circ = 138^\circ \)[/tex].

3. List all solutions within the given interval:
- Based on the periodic nature of the sine function and the interval [tex]\( 0^\circ < x < 360^\circ \)[/tex], the solutions are [tex]\( 42^\circ \)[/tex] and [tex]\( 138^\circ \)[/tex].

So, the values of [tex]\( x \)[/tex] for which [tex]\( \sin x = 0.6691 \)[/tex] in the interval [tex]\( 0^\circ < x < 360^\circ \)[/tex] are:

[tex]\[ x = 42^\circ \text { and } x = 138^\circ \][/tex]

These solutions are given to the nearest degree.