How do you solve $ {\sin ^2}x - 8\sin x - 4 = 0 $ and find all values of x in the interval $ \left[ {0,{{360}^0}} \right) $ ?

Hint: In order to determine the solution of the above trigonometric equation replace the $ \sin x $ as $ t $ . Compare the given quadratic equation with the standard form $ a{x^2} + bx + c $ to obtain the values for the variables. Now find the value of determinant using formula $ D = {b^2} - 4ac $ . You will get $ D > 0 $ so the roots are distinct and real. Obtain the roots using quadratic formula and put the value of $ t $ in the roots to obtain the required solution.

Complete step by step solution:
We are given a trigonometric equation $ {\sin ^2}x - 8\sin x - 4 = 0 $ and we have to find the solution in the interval $ \left[ {0,{{360}^0}} \right) $
Let $ t = \sin x $ . So substituting $ \sin x\,as\,t $ in the equation, we get
 $ {t^2} - 8t - 4 = 0 $
We have obtained a quadratic equation in $ t $ . Comparing the above quadratic equation with the standard quadratic equation $ a{x^2} + bx + c $ , we have
a=1
b=-8
c=-4
Let’s find the determinant of the above quadratic equation to understand the nature of roots by using the formula
 $ D = {b^2} - 4ac $
Putting the values of the variable , we get
 $
\Rightarrow D = {\left( { - 8} \right)^2} - 4\left( 1 \right)\left( { - 4} \right) \\
   = 64 + 16 \\
   = 80 \;
  $
Since, $ D > 0 $ then both the roots of the quadratic equation are distinct and real.
Root of the equation will be
 $
\Rightarrow {t_1} = \dfrac{{ - b}}{{2a}} + \dfrac{{\sqrt D }}{{2a}} = \dfrac{{ - \left( { - 8} \right)}}{2} + \dfrac{{\sqrt {80} }}{2} = 4 + \dfrac{{4\sqrt 5 }}{2} = 4 + 2\sqrt 5 \approx 8.472 \\
\Rightarrow {t_2} = \dfrac{{ - b}}{{2a}} - \dfrac{{\sqrt D }}{{2a}} = \dfrac{{ - \left( { - 8} \right)}}{2} - \dfrac{{\sqrt {80} }}{2} = 4 - \dfrac{{4\sqrt 5 }}{2} = 4 - 2\sqrt 5 \approx - 0.47 \;
  $
Value of $ {t_1} $ is rejected as the range of sine function is in the interval $ \left[ { - 1,1} \right] $ .
So we are left with
 $ {t_2} = - 0.47 $
Putting back the value of $ t $
 $
\Rightarrow \sin x = - 0.47 \\
\Rightarrow x = {\sin ^{ - 1}}\left( { - 0.47} \right) \;
  $
The value of x is an angle having sine value equal to $ - 0.47 $ . Since the value $ - 0.47 $ is not a remarkable value. Use the calculator to find the answer.
Sine function is negative in the 3rd and 4th quadrant . so we have two solution for the value of x
For the third quadrant we have
 $
\Rightarrow x = \pi + {\sin ^{ - 1}}\left( {0.47} \right) \\
\Rightarrow x = 180 + 28.16 \\
\Rightarrow x = {208.16^ \circ } \;
  $
And for the fourth quadrant, we have
 $
\Rightarrow x = 2\pi - {\sin ^{ - 1}}\left( {0.47} \right) \\
\Rightarrow x = 360 - 28.16 \\
\Rightarrow x = {331.84^ \circ } \;
  $
Therefore, the solution of the given trigonometric equation is $ x = {208.16^ \circ },{331.84^ \circ } $
So, the correct answer is “ $ x = {208.16^ \circ },{331.84^ \circ } $”.

Note: 1. The degree of the quadratic equation is of the order 2.
Every Quadratic equation has 2 roots.
Discriminant: $ D = {b^2} - 4ac $
Using Discriminant, we can find out the nature of the roots
If D is equal to zero, then both of the roots will be the same and real.
If D is a positive number then, both of the roots are real solutions.
If D is a negative number, then the root are the pair of complex solutions
Also remember,
1.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2. Period of sine function is $ 2\pi $ .
3. The domain of sine function is in the interval $ \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] $ and the range is in the interval $ \left[ { - 1,1} \right] $ .
4.Don’t forget to rearrange quadratic equations in the standard form.
5. Don’t forget to compare the given quadratic equation with the standard one every time.