Question

sin x = sin(x/2), evaluate x at range 0<=x<=2pi

Answer 1

i first use the double angle identites then i get:
2sin(x/2)cos(x/2) = sin (x/2)

then i change cos(x/2) to
\[\pm \sqrt{1 - \sin^{2}(\frac{ x }{ 2 })}\]

then i let y = sin (x/2), then i come to this expression

\[y^{2}(3-4y^{2}) = 0, y=0, \pm (\frac{ \sqrt{3} }{ 2 })\]

then my x = 0, 2pi, 2pi/3 and 4pi/3

but when i plug back x = 4pi / 3 in to the sin x = sin(x/2), i find that they re not equate, but instead they re in opposite sign but same magnitude. Anyone know why this approach is wrong?

Answer 2

you know you could have made this easier...

Answer 3

We have sin(x)=sin(x/2)
Set the insides equal.
Just to make sure we can find the most solutions.
Let's use the fact that sin(x)=cos(pi/2-x)
So cos(pi/2-x)=cos(pi/2-x/2)
Set insides equal.
Then let's use the fact cos is is even so cos(-x)=cos(x)
So cos(x-pi/2) =cos(pi/2-x/2)
set insides equal.
So we could also look at cos(pi/2-x)=cos(x/2-pi/2)
just make sure these x's are in the given interval
now let me take a look at your process.

Answer 4

oh wow i like the way you started

Answer 5

\[2\sin(x/2)\cos(x/2)-\sin(x/2)=0\]
now factor
\[\sin(x/2)(2\cos(x/2)-1)=0\]
set both factors equal to 0
\[\sin(x/2)=0 \] \[ 2\cos(x/2)-1=0\]

Answer 6

so 0<=x<=2pi
so 0/2<=x/2<=pi
let x/2=u
so we know that u is between 0 and pi including those endpoints.

sin(u)=0 when u=? on the interval [0,pi]
cos(u)=1/2 when u=? on the interval [0,pi]

this way right here is way better that writing the equation like 4 different ways like I did. lol.

Answer 7

than*

Answer 8

You should get 3 solutions from this.

Answer 9

Use the unit circle. Good state @hei .:)

Answer 10

ty

Answer 11

Hey but the reason your are going to have extra solutions (that are not actually solutions) is because you squared a radical.

Answer 12

Just like if we had
\[\sqrt{x+2}=-5\]
square both sides
x+2=25
x=23
this is not a solution

Answer 13

\[\sqrt{x^2-4x+10}=5 \]
square both sides
\[x^2-4x+10=25\]
\[x^2-4x-15=0 \]
we get \[x=\frac{4 \pm \sqrt{16-4(-15)}}{2}=\frac{4 \pm \sqrt{16+60}}{2}\]
But we have to plug back in to see if they both solutions because the square root functions had range [0, infinity) and not (-infinity, infinity)

Answer 14

Does this make sense?

Answer 15

Here is an easy problem to see what I mean:
\[\sqrt{x-1}=x-7\]
square both sides
\[x-1=(x-7)^2 \]
\[x-1=x^2-14x+49\]
\[0=x^2-15x+50\]
\[0=(x-5)(x-10) => x=5 \text{ or } x=10 \]
But we have to check since we got rid of even root.
\[\sqrt{10-1}=10-7 \text{ is true but } \sqrt{5-1} \neq 5-7 \]

Answer 16

so the only solution is x=10..
So since you got rid of even root in your problem you must go back and check your solutions.

Answer 17

I see, thanks again, a very good explanation