Question

Something isn't right with the way I've done this question. see attached file
6 sin (x) = 1 + 9 sin (x) solve over domain 0 x 2pi

Answer 1

What attached file?

Answer 2

here it is

Answer 3

1+9sin(x) -6sin(x) = 0
1 + 3sin(x) = 0
sin(x) = \(\sf \frac{1}{3}\)

Answer 4

therefore, principal value of x = arcsin(1/3) = 19.47º

Answer 5

negative or postive? Do I draw it in the first quadrant?

Answer 6

if we simplify that equation, we get:
\(3 \sin x=-1\)
therefore:
\[\sin x = - \frac{1}{3}\]

Answer 7

again we have this drawing: