Question

Given the following equations: Find (i) all solutions in the interval [0,2π ) in radians, and (ii) all solutions in radians:
(a) (2sin x − sqrt3)(sin x − 2) = 0

Answer 1

you guys dont have to give me all the answers, just the steps to get there

Answer 2

Wow, its been a long time since I did this. But since there's no one else around .... ;P

(2sin x − 3)(sin x − 2) = 0 so
(2sin x − 3) = 0 or (sin x − 2) = 0
2sin x = 3 sin x = 2
sin x = 3/2

Does that give you a start? 'Cause I'm really gonna have to dig (mentally) to figure out the next part. haha

Answer 3

lol i just realized i put the question wrong, its sqrt 3 not 3, does that change what you just wrote in anyway?

Answer 4

haha YES very much.
(2sin x − sqrt(3))(sin x − 2) = 0, right?

so 2sin x − sqrt(3) = 0
2sin x = sqrt(3)
sin x = sqrt(3)/2
(do you have a unit circle?) x = pi/3 or 4pi/3

Answer 5

oh shoot, wait that last one is NOT 4pi/3 but 2pi/3

Answer 6

yeah i was just about to say that lol since its positive, but thanks that helped me a lot for the first part.. now i just need to figure out what to do for part (ii)

Answer 7

well basically, all you need to do is add (or subtract) 2pi to each of the above answers continuously. Remember, 2pi = entire circle, so by adding or subtracting a circle or more, you can get all the solutions

Answer 8

i know the answer comes out to be x = pi/3 + 2pi(n), 2pi/3 + 2pi(n), (n) is an intiger

Answer 9

by the way, the sin x = 2 I got in my first post yields no solutions because sin x is true only in the interval [-1, 1]

Answer 10

yeah, that is the right answer. Do you understand it?

Answer 11

i was going to ask about the sinx = 2 and i see what you mean thank you, as for the answers, i understand the adding 2pi in order to get the solutions, but whats the (n) for? to show that you can add or subtract and amounts of circles?

Answer 12

any*

Answer 13

the (n) represents the number of circles or 2pi's

Answer 14

gotcha, would you mind running me through 1 more problem like that to see if i got it down? =P

Answer 15

sure np

Answer 16

2 sin^2 x + sinx-3 = 0

Answer 17

ok yeah, good start. ask if you need a hint

Answer 18

so 2sin^2x + sinx = 3 is the right first step? sorry

Answer 19

or 2sin^2x = 0 and sinx - 3 = 0?

Answer 20

2sin^2x + sinx = 3 is the right first step

Answer 21

okay yeah im lost again haha, would you mind running me through it? =P

Answer 22

haha ok. Well, this one's a little different. You gotta factor out the sin x to so you have parentheses (*ahem* factors) to work with

Answer 23

2sin^2x + sinx = 3
sin x (2 sin x + 1) = 3

Answer 24

so that becomes sinx(2sinx + 1) = 3? and sorry for late answers im getting ready for work lol, its in 5 mins

Answer 25

oh haha. yeah.
Well, just in case you want to know
xin x = 3 or 2 sin x + 1 = 3
not true 2 sin x = 2
sin x = 1

Answer 26

yeah.. sin x = 1 doesn't get you a very pretty answer on a calc, but that's as far as i remember

Answer 27

you're a boss lol thank you very much. i understand it a lot more than i did before hand much appreciated =)

Answer 28

haha glad i helped