Question

Find all solutions in the interval [0, 2π).

2 sin^2(x) = sin (x)

Answer 1

how do i do that?

Answer 2

what values of x would make sin(x) = 0?
Since (2) (0)^2 = 0 0= 0

Answer 3

?

Answer 4

wait

Answer 5

so would it be 0,pi,pi/6,5pi/6?

Answer 6

hello?

Answer 7

I'd say 0, pi, and 2pi

Answer 8

sin pi/6 = 0.5 and sin 5pi/6 = 0.5 as well.
Refer to circle of cos and sin here: http://fac-web.spsu.edu/math/edwards/1113/unit2.gif

Answer 9

ok can you help me with one more?

Answer 10

Sure

Answer 11

Find all solutions to the equation.

sin2x + sin x = 0
step by step would be fantastic

Answer 12

Just to make sure it's sin (2x) right?
not \[\sin ^{2} x\]

Answer 13

oh sorry its the second one
sin^2(x)

Answer 14

again, sin^2 x = (sin(x))^2
0+ 0 = 0, so x = 0, pi, 2pi are answers

Answer 15

there are more though

Answer 16

what are the others

Answer 17

(-1)^2 + (-1) = 0
so any sin(x) = -1 are answers. Find those values

Answer 18

how do i find those values

Answer 19

the circle I linked you.
(cos, sin) sin is the second value

Answer 20

it says 3pi/2

Answer 21

Yes, so x= 0, pi, 3pi/2, and 2pi

Answer 22

ok is there a referance video or something i can see to show me how to do that?

Answer 23

to do what? find sin(x) = certain answers or solving this problem?

Answer 24

solving this problem and others like it

Answer 25

Then my best suggestion is your textbook. The videos are really broad on trigonometry so I doubt you can find one that teaches you to solve certain problems like these. They cover more concepts.

Answer 26

If you have any questions, I'm here to help though. What part are you confused about?

Answer 27

alright thank you so much

Answer 28

Welcome :)