How do you solve $\sin^2 \theta = 2 \sin^2 \frac{\theta}{2}$ over the interval $[0,2pi]$ ?

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Answer 1 · 2014-11-20T04:34:07

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

by $sin^2theta=1-cos^2theta$ and $sin^2(theta/2)=1/2(1-costheta)$ ,

$=> 1-cos^2theta=1-cos theta$

by subtracting $1$ ,

$=> -cos^2theta=-cos theta$

by adding $cos theta$ ,

$=> cos theta-cos^2theta=0$

by factoring out $cos theta$ ,

$=> cos theta(1-cos theta)=0$

$=>{(cos theta=0 => theta= pi/2", "{3pi}/2),(cos theta=1 => theta=0", " 2pi):}$

Hence, $theta=0,pi/2,{3pi}/2,2pi$ .

I hope that this was helpful.

How do you solve \sin^2 \theta = 2 \sin^2 \frac{\theta}{2} \sin^2 \theta = 2 \sin^2 \frac{\theta}{2} over the interval [0,2pi][0,2pi]?