Answer to Question #6027 in Abstract Algebra for Samantha

Question

Find all solutions in the interval [0,2pi).
cos2x + sinx=1

Answer 1

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