Question 738948
use the half angle identities to find all solutions in the interval [0,2pi) sin^2x=cos^2(x/2)
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{{{sin^2(x)=cos^2(x/2)}}}
{{{cos(x/2)=sqrt((1+cos(x))/2)}}}
{{{cos^2(x/2)=(1+cos(x))/2}}}
{{{sin^2(x)=(1+cos(x))/2}}}
{{{2sin^2(x)=(1+cos(x))}}}
{{{2(1-cos^2(x))=(1+cos(x))}}}
{{{(2-2cos^2(x))=(1+cos(x))}}}
{{{(2cos^2(x))+cos(x)-1=0)}}}
{{{(2cos(x)-1)(cos(x)+1)=0}}}
..
2cos(x)-1=0
cos(x)=1/2
x=π/3, 5π/3
..
cos(x)+1=0
cos(x)=-1
x=π