Trig.
posted by mary .
Can someone please help me with this?
Find all solutions to the equation in the interval [0,2pi)
cos2x=sinx
I know I have to use some sort of identity, but I have no idea how to go about to solve this.

Trig. 
MathMate
Yes, by making use of the identity
cos2x = cos²xsin²x=12sin²x, the given equation can be transformed to:
2sin²x+sinx1=0
which is a quadratic in sin(x).
Solving the quadratic by factoring,
(2sinx1)(sinx+1)=0
or sin(x)=1/2 or sin(x)=1
Find all roots in the interval [0,2π]. 
Trig. 
mary
Thank you very much for the help

Trig. :) 
MathMate
You are welcome!
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