How do you solve $sin^2 (theta) = 1/2$ ?

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Answer 1 · 2016-02-11T22:55:41

$theta= pi/4 +(kpi)/2$

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

$sin(theta)=1/sqrt(2)=sqrt(2)/2 or sin(theta)=-1/sqrt(2)=-sqrt(2)/2$

$theta=pi/4+2kpi or theta=3pi/4+2kpi or$

$theta=5pi/4+2kpi or theta=7pi/4+2kpi$

If you trace these four solutions in the trigonometrical circle you will see that the four can be resumed in just one:

$theta= pi/4 +(kpi)/2$

How do you solve sin^2 (theta) = 1/2 sin^2 (theta) = 1/2 ?