What is the answer for 2sin^2(theta/4)=12sin2(θ4)=1?

I understand the answer key up to sin(theta/4)=-1/sqrt(2)sin(θ4)=12.
Then, the next step is
theta/4=(5pi)/4θ4=5π4.
How does that work?

1 Answer
Aug 2, 2017

The solutions are S={pi+8pin, 3pi+8pin, 5pi+8pin, 7pi+8pin }S={π+8πn,3π+8πn,5π+8πn,7π+8πn}, AA n in ZZ

Explanation:

The equation is

2sin^2(theta/4)=1

sin^2(theta/4)=1/2

sin(theta/4)=+-1/sqrt2

If,

sin(theta/4)=1/sqrt2, =>, theta/4=pi/4+2pin and theta/4=3/4pi+2pin, AA n in ZZ

Therefore,

theta=pi+8pin and theta=3pi+8pin, AA n in ZZ

sin(theta/4)=-1/sqrt2, =>, theta/4=5/4pi+2pin and theta/4=7/4pi+2pin, AA n in ZZ

Therefore,

theta=5pi+8pin and theta=7pi+8pin, AA n in ZZ