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

Question

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

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Answer 1 · 2017-08-02T11:48:21

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

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#