Question

find the values of x in the interval [x,2pi] that satisfy

sin^2x + sinx - 2 = 0

Answer 1

(sin(x) +2)(sin(x)-1)

Answer 2

how did you get that

Answer 3

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

Answer 4

Just factor it.
How do you factor \[u^2 +u-2=(u+2)(u-1)\]

Answer 5

Since sin(x) cannot equal 2
then
sin(x) =1
hence
\[
x=\frac \pi 2 + 2\ k \pi
\]

Answer 6

Did you get it?

Answer 7

im lost at sinx not = to 2

Answer 8

Sin x varies between 0 and 1
Solutions in interval [x,2pi] required by the question

Answer 9

So
\[
x=\frac \pi 2
\]

Answer 10

what does k stand for?