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

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

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Answer 1 · 2014-11-20T04:34:07

${sin}^{2} θ = 2 {sin}^{2} (\frac{θ}{2})$ $sin^2theta=2sin^2(theta/2)$

by ${sin}^{2} θ = 1 - {cos}^{2} θ$ $sin^2theta=1-cos^2theta$ and ${sin}^{2} (\frac{θ}{2}) = \frac{1}{2} (1 - cos θ)$ $sin^2(theta/2)=1/2(1-costheta)$ ,

$\Rightarrow 1 - {cos}^{2} θ = 1 - cos θ$ $=> 1-cos^2theta=1-cos theta$

by subtracting $1$ $1$ ,

$\Rightarrow - {cos}^{2} θ = - cos θ$ $=> -cos^2theta=-cos theta$

by adding $cos θ$ $cos theta$ ,

$\Rightarrow cos θ - {cos}^{2} θ = 0$ $=> cos theta-cos^2theta=0$

by factoring out $cos θ$ $cos theta$ ,

$\Rightarrow cos θ (1 - cos θ) = 0$ $=> cos theta(1-cos theta)=0$

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

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

I hope that this was helpful.

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

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