Hi Steven434, welcome to the forum :)
This is NOT a calculus question - it is a trig question ://
How would you solve this trig equation?
2sin2x+sinx=0
Where 0<x<360
\( 2sin2x+sinx=0 \\ 2(sin^2x-cos^2x)+sinx=0\\ 2(sin^2x-[1-sin^2x])+sinx=0\\ 2(sin^2x-1+sin^2x)+sinx=0\\ 2(2sin^2x-1)+sinx=0\\ 4sin^2x-2+sinx=0\\ 4sin^2x+sinx-2=0\\ let\;y=sinx\\ 4y^2+y-2=0 \)
4y^2+y-2=0 = { y=-(((sqrt(33)+1)/8)), y=((sqrt(33)-1)/8)}
-(((sqrt(33)+1)/8)) = -0.8430703308172536
((sqrt(33)-1)/8) = 0.5930703308172536
so
sinx = - 0.8430703308172536 (this will be in 3rd and 4th quadrants)
OR sinx = 0.5930703308172536 (this will be in 1st and 2th quadrants)
sinx = 0.8430703308172536 OR sinx = 0.5930703308172536
x=asin(0.8430703308172536)
asin(0.8430703308172536) = 57.46577 degrees
asin(-0.8430703308172536) = 180+57.46577 or 360-57.46577 degrees
asin(-0.8430703308172536) = 237.46577 or 302.53423 degrees
asin(0.5930703308172536) = 36.375192227 or 180-36.375192227
asin(0.5930703308172536) = 36.375192227 or 143.6248077 degrees
So to 2 decimal places the answers are
36.38 and 143.62 and 237.47 and 302.53 degrees