# Trig.

posted by mary

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.

1. MathMate

Yes, by making use of the identity
cos2x = cos²x-sin²x=1-2sin²x, the given equation can be transformed to:
2sin²x+sinx-1=0
which is a quadratic in sin(x).
(2sinx-1)(sinx+1)=0
or sin(x)=1/2 or sin(x)=-1
Find all roots in the interval [0,2π].

2. mary

Thank you very much for the help

3. MathMate

You are welcome!

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