# Answer to Question #6027 in Abstract Algebra for Samantha

Question #6027

Find all solutions in the interval [0,2pi).

cos2x + sinx=1

cos2x + sinx=1

Expert's answer

we know that 1-cos(2x)=2sin^2(x) so cos(2x)+sin(x)=1 we can rewrite as

sin(x)=2sin^2(x) so sin(x)=0 or sin(x)=1/2 and we have from first x=0, x=pi

from second x=pi/6 , x=5*pi/6

sin(x)=2sin^2(x) so sin(x)=0 or sin(x)=1/2 and we have from first x=0, x=pi

from second x=pi/6 , x=5*pi/6

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