how do i solve the following equation for values of x in the range
0° < x < 360°
6sin²x - sinx - 2 = 0
Let w = sinx
Then w² = sin²x
6w² - w - 2 = 0
Factor as
(2w + 1)(3w - 2) = 0
2w + 1 = 0; 3w - 2 = 0
2w = -1; 3w = 2
w = ; w =
Now replace w by sinx
sinx = ; sinx =
The first one can be solved as special angles.
The sine is negative in the 3rd and 4th quadrants,
and since the reference angle 30° has sine +,
the angles in the 3rd and 4th quadrants having that
sine are 210° and 330°.
The second one can't be solved as a special angles.
The sine is positive in the 1st and 2nd quadrants,
We find the refrence angle using inverse sine on a
calculator as 41.8103149°. So the angles in the 1st
and 2nd quadrants having that sine are 41.8103149°
for the 1st quadrant answer and to find the 2nd
quadrant answer we subtract 41.8103149° from 180°.
180° - 41.8103149° = 138.1896861°
The four solutions are:
x = 210°, 330°, 41.8103149°, and 138.1896861°
Edwin