SOLUTION: Solve the equation for all solutions in the interval 0degree &#8804; x < 360degree. 2sin^2 x-sinx-1=0

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Question 661718: Solve the equation for all solutions in the interval 0degree ≤ x < 360degree.
2sin^2 x-sinx-1=0
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
If you understand how to solve this algebraic equation for y:  

                  2y² - y - 1 = 0
              (2y + 1)(y - 1) = 0

            2y + 1 =  0;   y - 1 = 0
                2y = -1;       y = 1
                 y = -1%2F2  


then you can understand how to solve this trigonometric equation
for sin(x):


         2sin²(x) - sin(x) - 1 = 0              

    [2sin(x) + 1][sin(x) - 1] = 0

   2sin(x) + 1 = 0;     sin(x) - 1 = 0

      2sin(x) = -1;         sin(x) = 1

       sin(x) = -1%2F2

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To solve

sin(x) = -1%2F2

2e remember that sin(30°) = 1%2F2 then 
you know that 30° is the reference angle.  The
fact that the sine is a negative number, that
tell you that the angle x is in the 3rd or
4th quadrant.

To get the angle in the 3rd quadrant which has 
referent angle 30², we add 30² to 180° and get
210°.

To get the angle in the 4th quadrant which has 
referent angle 30², we subtract 30° from 360° 
and get 330°.

So two of the solutions are 210° and 330°. 

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To solve sin(x) = 1,

we remember that sin(90°) = 1

So the other solution is 90°.

The three solutions are then:

x = 90°, x = 210°, and x = 330°

Edwin