Find the solution in quadrant III for 40 + 55sin(15x) = 1.1. θ<0

Question

Find the solution in quadrant III for 40 + 55sin(15x) = 1.1. θ<0

Answer 1

So we have sin(15x) = (1.1-40)/55 = - 38.9/50

The reference angle would be sin^-1(38.9/50)

In III quadrant it would be sin^-1(38.9/50) + π

Or if you need it negative - than - π

So now you have 15x = sin^-1(38.9/50) + π + 2πk

and

x = 1/15 (sin^-1(38.9/50) + π) + 2πk/15

Answer 2

 40 + 55sin(15x) = 1.1

-40 -40

___________________________________

55sin(15x) = -38.9

" " -.7072......

sin x -.04715... =-2.7025...°