Question 808521
Given:
(1) {{{sin^2(x) - 12cosx -12 = 0}}}
Use the trig identity
(2) {{{sin^2(x) + cos^2(x) = 1}}} to get
(3) {{{sin^2(x) = 1 - cos^2(x)}}} 
Put (3) into (1) to obtain
(4) {{{1 - cos^2(x) - 12cosx -12 = 0}}}   or
(5) {{{- cos^2(x) - 12cosx - 11 = 0}}}  or
(6) {{{cos^2(x) + 12cosx + 11 = 0}}}  
which factors into
(7){{{(cos(x) + 1)*(cosx + 11) = 0}}} 
Now set each factor equal to zero and solve for x, and get
(8){{{cos(x) + 11 = 0}}} and
(9){{{cosx + 1 = 0}}} or
(10){{{cosx = -11}}}     and
(11){{{cosx = -1}}}
Since absolute value of cos(x) is never greater than one, (10) has no solutions, whereas (12) is
(13) x = arccos(-1) or
(14) x = 180 degrees
Let's check this with (1).
Is (sin(180)^2 -12cos(180) -12 = 0)?
Is (0^2 - 12*(-1) - 12 = 0)?
Is (0 + 12 - 12 = 0)?
Is (0 = 0)? Yes
Answer: the only x in the interval [0,360] is x = 180 degrees or {{{pi}}} radians.