Question

Find all values of x such that \(\sin 2x = \sin x\) and 0 < x < 2\(\pi\)

Answer 1

i suppose the only way for this to happen is if the x were 0....

Answer 2

yeah or pi, but you specified x < 2pi

Answer 3

hmm i suppose pi works

Answer 4

sin 2x = sin x

arcsin both sides

2x = x

2x - x = 0

x = 0

how can i get the other values?

Answer 5

hmmm good question

Answer 6

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

Therefore,

sin(2x)=sin(x) -->
2sin(x)cos(x)=sin(x)
2sin(x)cos(x)-sin(x)=0
sin(x)[2cos(x)-1]=0

Now either

sin(x)=0 ---> x=0,pi,...

or

2cos(x)-1=0 --> cos(x)=1/2 ---> x=pi/3

Answer 7

then i suppose the others can be solved by adding pi?

Answer 8

Why do you want to add a pi?

Answer 9

because one angle is missing

Answer 10

You have only 3 solutions

Answer 11

0, pi/3, pi

Answer 12

there's actually 5.. but i know how to get one of the missing angles...

Answer 13

oh ya i though x between 0 and pi

Answer 14

sure 2pi is the 4th solution

Answer 15

yes

Answer 16

and x=2pi-p/3=5pi/3

Answer 17

is the 5th solution

Answer 18

oh subtract pi/3 from 2pi...ah yes... the negative angle

Answer 19

exactly ...since cos(x) is positive in the 4th quad

Answer 20

Here's an example you might want to look at:

tan(4x) = -tan(2x) 0° ≤ x ≤ 360°
tan(4x) = tan(-2x)

****DONT DO THIS: 4x = -2x
INSTEAD... 4x = 180n -2x (Write one side as a general solution)

then simplify:

6x = 180n
x = 30n

then you have all possible values of x
now find all the ones that work, and you have all solutions