Question

Can someone help me understand this?
Find all solutions in the interval [0, 2π).

(sin x)(cos x) = 0

Answer 1

do u understand the interval notation [ )

Answer 2

u would think where is sinx = 0 cosx = 0 at around 2pi

Answer 3

No, I don't. The way my online class explains it is really hard to grasp. I was hoping someone knew a simpler way.

Answer 4

i think one of them would be pi/2 cuz when sin(pi/2)*cos(pi/2) = 0 holds true

Answer 5

basically 90 180 270 degrees will hold true

Answer 6

which is pi/2 , pi, 3pi/2

Answer 7

well there are lots of ways to view this

where does

sin(x)= 0

and where does

cos(x) = 0

if either factor is zero, then the product will be zero...

and alternative is to use substitution

sin(2x) = 2sin(x)cos(x)

so you can say

1/2sin(2x) = 0

hope it helps

Answer 8

u don't want to go to 2pi or more cuz that would include outside ur interval notation

Answer 9

0 is also an answer

Answer 10

How did you do that substitution? Where did you pull the 2x from?

Answer 11

well there is a thing called double angle trig identities

one of which is

sin(2x) = 2sin(x)cos(x)

if you haven't covered it... just look at where

cos(x) = 0 and sin(x) = 0

Answer 12

They both =0 at different times.

Answer 13

thats correct...

because you are multiplying

sin(x)cos(x) = 0

if

sin(x) = 0...... then the product of the 2 terms is zero

and

cos(x) = 0 ..... then the product of the 2 terms is zero

thats why there well be 4 solutions.....you're interval doesn't include 2pi...

which would have been a 5th solution.

Answer 14

So they could be.. pi, 3pi/2, and pi/2... Would zero count?

Answer 15

yes... since the interval says state at 0 and go up to 2pi... but don't include it

Answer 16

Oh, okay. So my answer will only have those three?

Answer 17

Uhmm.. None of my choices have three..

Answer 18

when u have [ ] it includes those in the interval when u have ( ) these are open and don't include in the interval so u have [ )

Answer 19

thats because there are 4....

Answer 20

u have 0 , pi/2, pi, and 3pi/2

Answer 21

Oh! SO because the zero has a [, it's included, but not 2pi, because it has a )?

Answer 22

thats correct

Answer 23

Is this one the same way?
Find all solutions in the interval [0, 2π).

2 sin^2x = sin x

Answer 24

well for that subtract sin(x) from both sides of the equation

then factor sin(x)

sin(x)(2sin(x) - 1) = 0

now you solve each factor to find the solutions.

Answer 25

How did he get that from subtracting sin x?

Answer 26

@whpalmer4

Answer 27

took out a sinx as above was written 2sinx^2 - sinx =0
so sinx(2sinx - 1) = 0

Answer 28

What do I do next? Graph them, and see where they equal zero? @fiberdust

Answer 29

if you have

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

-sin(x) -sin(x)
----------------------

2sin^2(x) - sin(x) = 0

now factor

sin(x)(2sin(x) - 1) = 0

so you need to find by solving where

sin(x) = 0

and

2sin(x) - 1 = 0

hope it makes sense.