Question

Solve on the interval [0,2pi)
2sin^2x+3sinx+1=0

Answer 1

@mathmate @satellite73 @freckles

Answer 2

factor

Answer 3

set each factor equal to zero, solve for sine
then solve for x

Answer 4

let sin(x) = y \[2y^2+3y+1=0\] can you factor the quadratic equation
?

Answer 5

well i tried before but i can try again

Answer 6

nice try it let me know what youge :=))

Answer 7

t*

Answer 8

x=-3+sqrt1/4
x=-3-sqrt1/4

Answer 9

@Nnesha

Answer 10

do you mean \[x=\frac{ -3 \pm \sqrt{1} }{ 4 }\] ?

Answer 11

Yeah i just wrote them out, same thing

Answer 12

i just want make sure it's (-3-sqrt{1}) over 4
or -3 - (sqrt{1}/4})

Answer 13

alright square rooot of 1 is just one \[\frac{ -3 +1 }{ 4 }~,~~\frac{-3-1}{4}\] simplify this brb let me login to other laptop

Answer 14

okay

Answer 15

hmm there is a typo

Answer 16

x=-1, -1/2

Answer 17

right

Answer 18

yes that's right \[\frac{ -3+1 }{ 4 } \rightarrow \frac{-2}{4} \rightarrow \frac{-1}{2}\]

Answer 19

sweet but how arethese answers relevant to the solution because i know that these arent the answers since my assignment has answers like 2pi/3

Answer 20

so sin(x) = -1/2 and -1
u can use unit circle to find exact solution both values solutions are negative
sign chart
so answer would be in 3rd and 4th quadrant