What is the answer for $2 {sin}^{2} (\frac{θ}{4}) = 1$ $2sin^2(theta/4)=1$ ?

Question

I understand the answer key up to $sin (\frac{θ}{4}) = - \frac{1}{\sqrt{2}}$ $sin(theta/4)=-1/sqrt(2)$ .
Then, the next step is
$\frac{θ}{4} = \frac{5 π}{4}$ $theta/4=(5pi)/4$ .
How does that work?

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Answer 1 · 2017-08-02T11:48:21

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

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