Find all the solutions in the interval [0, 2pi): 5 sin^2 x + 8 sin x - 4= 0?

Question

a. 0.4115, 2.7301
b. pi/6, 5pi/6
c. 0.4, 2.7416
d. 3.5531, 5.8717
e. none of these

Answer 1

here's a neat trick to use
let x=sinx
then you have 5x^2+8x-4=0
(5x-2)(x+2)
then get x by itself
x=2/5
x=-2
so when does sin x=2/5
sin x=-2?

Answer 2

5 sin^2 x + 8 sin x - 4= 0
sin(x) = (- 8 ± √(64 + 80))/10
sin(x) = (- 8 ± √144)/10
sin(x) = (- 8 ± 12)/10
sin(x) = (- 8 - 12)/10, (- 8 + 12)/10
sin(x) = (- 20/10, 4/10)
-2 is not possible, so
x = sin^-1(0.4)
x = 0.4115168, 2.730076