When will (-1/4)sin(x/2) = 0?

Question

I know one answer is when x=0.

I know the other 2 possibilties are 2pi and 4pi, but how do I get those answers? I've looked at the trig circle, but it's not making much sense.

Oh, I just figured something out: sin180 and sin360 give me 0. So, according to the trig circle, I use the y-value for sin. Is it because (a,b) is (cos, sin) on the trig circle?

Answer 1

When an angle = 0 or 180 degrees in the unit circle, the value for sine = 0. For your equation x/2 must equation 0 or 180 for the sine of (x/2) to equal 0, so x = 0 and 360. (multiplying zero by -1/4 has no effect).

In radians this is equivalent to 0 and 2pi. (and any even multiple of pi since you can rotate about the unit circle a full revolution and get the same value for sine)

Answer 2

math_kp is correct
another way to solve this
is to consider
1) arcsin(0)=x/2 is given
and
2) arcsin(0)=theta
which is the same as
3) theta=arcsin(0)
and
4) theta=x/2
we know that is at
theta = 0,pi,2pi,3pi
or theta=npi
so use 4)
npi=x/2
therefore
2npi=x
shows mathematically you are correct
in your observation of the unit circle
you can double check
(-1/4)sin(2npi/2)=
(-1/4)sin(npi)=
(-1/4)(0)=0
yep that's it

Answer 3

sin(x/2)=0 means that x/2 =k*pi (k any integer)
so x= 2k*pi which means an integer number of turns(2pi ia a complete turn) so you end always at the same point of the trig.circle.

Answer 4

above is true when (deviding by (-1/4)

sin (x/2) = 0

general solution is x/2 = n*pi as sin(npi) =0

or x= 2n *pi