Question

solve: 2sin²θ-sinθ=0
please show how you got your answer
Thank you so much!!

Answer 1

if the problem were

2 x^2 -x=0
could you find the values for x ?

Answer 2

i got down to:
2cosθ=1 and cosθ=1/2

Answer 3

you mean sin , don't you ?

but can you solve
2 x^2 -x=0
?

Answer 4

yeah sorry

Answer 5

what do you get ?

Answer 6

oh i forgot to add in the equation "-1" after the sin

Answer 7

what do you get ?

Answer 8

I would factor out an x from both terms as a start.

Answer 9

okay so it would be 2x-1, right?

Answer 10

or are you saying we have a different problem than what you posted ?

Answer 11

and then x=0 and x=1/2

Answer 12

the problem is: 2sin²θ-sinθ-1=0

Answer 13

ok, so we want to solve
2x^2 -x -1 =0

If it factors it will be
(2x 1 )(x 1 )=0
and we need to put in the correct signs
the -1 at the end means the signs will be different
the -x means the bigger one is minus
can you figure it out ?

Answer 14

so it will be (x-1) and (x+.5)
which then will be x=1 and x=-.5

Answer 15

and then i got cosθ=1 and cosθ=-.5 is this right, i tried using u-substitution because my teacher told me to

Answer 16

if the problem is
2sin²θ-sinθ-1=0

then x is really sin of some angle so you have
sin θ = 1 or sin θ = -0.5

I am not sure why you think it is cosine ??

Answer 17

sorry i just got confused for a sec, but i know its sin

Answer 18

one way to find the values of sin is plot it from 0 to 2pi (or 0 to 360 degrees)