Answer :
This all I got to help you.....
cos half-angle formula:
cos x/2=√[(1+cosx)/2]
√[(1+cosx)/2]-sinx=0
√[(1+cosx)/2]=sinx
square both sides
(1+cosx)/2=sin^2x=1-cos^2x
1+cosx=2-2cos^2x
2cos^2x+cosx-1=0
(2cosx-1)(cosx+1)=0
..
2cosx-1=0
cosx=1/2
x=π/3 and 5π/3
or
cosx+1=0
cosx=-1
x=π
HOPE I HELPED VANESSAN.. I HOPE ITS NOT THAT CONFUSING TOWARDS YOU :D
cos half-angle formula:
cos x/2=√[(1+cosx)/2]
√[(1+cosx)/2]-sinx=0
√[(1+cosx)/2]=sinx
square both sides
(1+cosx)/2=sin^2x=1-cos^2x
1+cosx=2-2cos^2x
2cos^2x+cosx-1=0
(2cosx-1)(cosx+1)=0
..
2cosx-1=0
cosx=1/2
x=π/3 and 5π/3
or
cosx+1=0
cosx=-1
x=π
HOPE I HELPED VANESSAN.. I HOPE ITS NOT THAT CONFUSING TOWARDS YOU :D
