Question

plzzzz help: find all values of theta in the interval 0 12 years ago

Answer 1

3cos2x + 2sinx + 1 = 0

use the relation cos2x = 1 - 2 sin^2x
to get an equation is sinx

Answer 2

wait is tht an identity? cos2x = 1 - 2 sin^2x

Answer 3

yes

Answer 4

so replace cos3x by 1 - 2sin^2 x and you'll get a quadratic equation in sin x

Answer 5

* replace cos 2x

Answer 6

so um cos2x=3cos2theta?

Answer 7

oh yes i was lazy so i put x instead of theta

Answer 8

what do i do wit the rest? 2sin theta+1=0?

Answer 9

ok ill write everything in theta

cos 2 theta = 1 - 2sin^2 theta

so 3(1 - 2sin^2 theta) + 2 sin theta + 1 = 0

now simplify this to get a quadratic in sin theta

Answer 10

i got sin^2theta=1

Answer 11

correct so faR?

Answer 12

@cwrw238

Answer 13

i got theata=90?

Answer 14

hold on

Answer 15

okk

Answer 16

3(1 - 2sin^2 theta) + 2 sin theta + 1 = 0
3 - 6sin^2 theta + 2 sin theta + 1 = 0
6sin^2 theta - 2 sin theta - 4 = 0
3 sin^2 theta - sin theta - 2 = 0

Answer 17

did you get that?

Answer 18

factoring
(3sin theta + 2)(sin theta - 1) = 0

Answer 19

oh isee wht i did. one sec plz

Answer 20

y -4? its shuld be positive 3+1=4

Answer 21

and y r u halfing the 6sin^2theta nd everything else?

Answer 22

so -3 on both sides udid?

Answer 23

ok so now i have 3-6sin^2theta+2sin theta+1=0 ---->do i subtract 3 on both sides?

Answer 24

need the slope

Answer 25

wait 1-3 is -2

Answer 26

how did u gt -4?

Answer 27

-1 -3 = -4

Answer 28

wrong:3-6sin^2theta+2sin theta+1=0 is this right:3-6sin^2theta+2sin theta-1=0 i had +1 insteads of -1

Answer 29

howcom ur doin tht? i nevr did be4

Answer 30

if you like you can look at it this way
I multiplied each term by -1
in order that first term (the 6 sin^2 theta becomes positive
the equation still hodls and is easier to factor

Answer 31

but my way wont work?

Answer 32

what was your way?