Question

Find all solutions of the equation on the intervals [0,2pi).

2sin^2 x+2 sin x-3=0

Answer 1

what math are you in?

Answer 2

per-calculs

Answer 3

oh i'm only in Algebra 1 sorry :/ maybe @Notamathgenius can help you tho :)

Answer 4

(don't let the name trick you ) lol

Answer 5

@ciecey ??
@Notamathgenius ?

Answer 6

substituting z for sin(x), you get:
z^2 - 2z - 3 = 0
(z-3)(z+1)=0
so z, or sin(x), must either equal 3 or -1

the range of sin(x) is -1 to 1 so 3 is out of the question.

Thus, setting sin(x) = -1, you get 3pi/2 radians, or 270 degree

Answer 7

thank u @Notamathgenius

Answer 8

You're welcome

Answer 9

good job :D)