Solve for theta 2Sin^2 + Sin - 1=0?

Question

O < theta < 360

Answer 1

Let x = sin theta.
Then 2x^2 + x -1 = 0.
(2x - 1)(x+1) = 0
x = 1/2 or x = -1.
If x = 1/2 then theta = 30 deg. or 150 deg.
If x = -1 then theta = 270 deg.

Answer 2

Let y = sine of theta.
Then your equation is 2y^2 + y - 1 = 0.
This is just a quadratic equation where a=2, b = 1, and c = -1.
Solve it using the quadratic formula and get

y = (-1 - 3) /4 or y = (-1 + 3) / 4 , that is, y = -1 or y = 1/2

But remember that y = sine of theta, so if sine of theta = -1 then theta = 270 is the only possible value for 0 < theta < 360.

If sine of theta = 1/2, then theta = 30 or theta = 150.

Your equation has three solutions : theta = 30, theta = 150, or theta = 270

Answer 3

2sin^2x+sinx-1=0
2sin^2x+2sinx-sinx-1=0
2sinx(sinx+1)-1(sinx+1)=0
(sinx+1)(2sinx-1)=0
sinx=-1 or 1/2
if sinx=-1 then x=270*
if sinx=1/2,x=30* or 150*