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.