Question

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

sin2x + sin x = 0

Answer 1

Is it sin2x or sin^2 x? :D

Answer 2

sam question here :))

Answer 3

same*

Answer 4

sin^(2)x

Answer 5

\[\sin^2 x + \sin x = 0 \]

\[sinx(sinx+1) = 0\]

\[Thus=> sinx = 0,\sin x =-1\]

So x can be 1' or 180'

Understood right? :)

Answer 6

ok

Answer 7

Understood? Yes/No ?

Answer 8

yes

Answer 9

Awesome!! :D

Answer 10

wahts the answer though

Answer 11

@AkashdeepDeb come back pl

Answer 12

btw the answer is wrong.

x is ± k*pi , where k is any integer
or
x is 3pi/2 ± 2*k*pi

Answer 13

x = 0, π, four pi divided by three, five pi divided by three

x = 0, π, pi divided by three,, two pi divided by three

x = 0, π, five pi divided by three, five pi divided by three

x = 0, π, three pi divided by two

Answer 14

answer choices

Answer 15

last choice: d

Answer 16

r u shure

Answer 17

i thought it was the second

Answer 18

based on?

lol

yes. i'm sure.