How do you solve #sin^2 x- 8 sin x - 4= 0# and find all values of x in the interval# [0, 360^o)#?

#f(x) = sin^2 x - 8sin x - 4 = 0#. Call #sin x = t#, we have:
#f(t) = t^2 - 8t - 4#. This is a quadratic equation. Solve it by using the new quadratic formula #d^2 = b^2 - 4ac = 64 + 16 = 80 = 16*5 rArr d = 4*sqrt(5)#.
#x_1 = -b/(2a) + d/(2a) = 8/2 + (4*sqrt(5))/2 = 4 + 2sqrt(5) approx 8.472# (rejected)
#x_2 = -b/(2a) - d/(2a) = 4 - 2sqrt(5) approx -0.47#
Next, solve #t = sin x = -0.472#. Calculator give #x_1 = 180 + 28.16 = 208.16°#.
Trig unit circle gives another arc #x# that has the same sine value (#-0.472#).
#x_2 = 360 - 28.16 = 331.84°#
Answers within interval #(0°, 360°)#: #x_1 = 208.16°#; #x_2 = 331.84°#.
Check
#x = 208.16° rArr sin x = -0.472 rArr f(x) = (-0.472)^2 - 8(-0.472) - 4 = 0.22 + 3.78 - 4 = 0# Correct.

I propose a slightly (but not too) different way of obtaining the solution (approximation in the very last step).

We start from #sin^2 x−8 sin x−4=0# and the first step, as in Nghi's answer, is to substitute #t=sin x#. We get #t^2-8t-4=0#, which is a quadratic formula whose solutions are

#t_1=frac{-(-8)+sqrt{(-8)^2-4*1*(-4)}}{2*1}=4+2sqrt{5}#
#t_2=frac{-(-8)-sqrt{(-8)^2-4*1*(-4)}}{2*1}=4-2sqrt{5}#

To get the solution of the initial equation, we have to find all #x_1 in [0°,360°)# such that #sin x_1=4+2sqrt{5}# and all #x_2 in [0°,360°)# such that #sin x_2 =4-2sqrt{5}#.

It turns out that #sin x_1=4+2sqrt{5}# has no solution #x_1#, since #sin x_1 in [-1,1]# for all #x_1 in [0°,360°)# and #4+2sqrt{5}>1#.

On the other hand, #sin x_2 =4-2sqrt{5}# has at least a solution: it's easy to check that #-1<4-2sqrt{5}<0#.
graph{sin(x) [-0.1, 6.4, -1.1, 1.1]}
The graph of the sine function suggests that the solutions will be two, both in the interval #[180°,360°)#.

So we get
#x_{2,1}=180°-arcsin(4-2sqrt{5})#
#x_{2,2}=360°+arcsin(4-2sqrt{5})#

Since #arcsin(4-2sqrt{5}) approx -28.173°# we get that #x_{2,1} approx 208.173 °#and #x_{2,2} approx 331.827 °#.