solve this pre cal equation?

Question

sin theta = cos (theta/2) showing every step

Answer 1

2sin thete/2 cos theta/2=cos theta/2
2sintheta/2cos theta/2-cos theta/2=0
cos theta/2(2sintheta/2-1)=0
cos theta/2=0=pi/2
theta/2=2npi-pi/2
theta=(4n-1)pi
2sin theta/2-1=0
2sin theta/2=1
sin theta/2=1/2=pi/6
theta/2=npi+(-1)^n*(pi/6)
theta=2npi+(-1)^6*pi/3

Answer 2

sinz=cosz/2 [I have put z=theta]
2sinz/2*cosz/2=cosz/2
2 sinz/2=1
sinz/2=1/2
z/2=30deg
z=60deg
[please note I have taken the simple solution for z in the first quadrant. You can have several solutions by adding 360deg.]
z=Theta=60deg

Answer 3

YOUR QUESTION IS NOT AN EQUATION BUT A UNIQUITY.ASSUMING IT TO BE TRUE ,LET US FIND THE VALUE OF @ WHICH SATISFIES IT
L.H.S=SIN@
=COS(90-@)
R.H.S=COS(@/2)
ON COMPARISON WE GET 90-@=@/2
90=@+(@/2)
90=(3@)/2
180=3@
180/3=@
@=60


THUS YOUR EQUATION HOLDS ONLY FOR @=60 AND NOT FOR ALL @

Answer 4

set u = theta/2
sin(2u) = cos(u)
*trig identity
2sin(u)cos(u) = cos(u)
*divide by cos(u)
2sin(u) = 1
*divide by 2
sin(u) = 1/2
u = 30 degrees or pi/6
*therefore
**********
theta = 60 degrees or pi/3