How do you solve $sin^2 theta = 3/4$ between 0 and 2pi?

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Answer 1 · 2016-03-10T09:48:45

$theta=pi/3, (2pi)/3, (4pi)/3, (5pi)/3$

$sin^2theta=3/4hArr(sin^2theta-3/4)=0$ or

$(sin^2theta-(sqrt3/2)^2)=0$ or

$(sintheta-sqrt3/2)(sintheta+sqrt3/2)=0$

i.e. $sintheta=+-sqrt3/2$

Hence, $theta=pi/3, (2pi)/3, (4pi)/3, (5pi)/3$

How do you solve sin^2 theta = 3/4sin^2 theta = 3/4 between 0 and 2pi?