How do you solve for all possible solutions between 0 and 360?
(sin(x)^2)-(12sin(x))+2=0
+118723 I am just going to outline the answer. If you have questions then ask. 
Let y=sinx where $$-1\le y\le1$$
$$y^2-12y+2=0$$
solve using the quadratic formula.
I got 2 answers but only one is within the limits
y= -0.169048
sinx=-0.169048
sin is neg so x is in the 3rd or 4th quads
Find the 1st quad equivalent
inverse sin 0.169048 = 9.73247 degrees
3rd quad angle x= 180+9.73248 = 189.73248 = 189°43'57"
4th quad angle x= 360-9.73248 = 350.26752 = 350°16'3"
+118723 I am just going to outline the answer. If you have questions then ask.
Let y=sinx where $$-1\le y\le1$$
$$y^2-12y+2=0$$
solve using the quadratic formula.
I got 2 answers but only one is within the limits
y= -0.169048
sinx=-0.169048
sin is neg so x is in the 3rd or 4th quads
Find the 1st quad equivalent
inverse sin 0.169048 = 9.73247 degrees
3rd quad angle x= 180+9.73248 = 189.73248 = 189°43'57"
4th quad angle x= 360-9.73248 = 350.26752 = 350°16'3"
