Posted by **mary** on Saturday, February 27, 2010 at 8:55pm.

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**, Saturday, February 27, 2010 at 9:25pm
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).

Solving the quadratic by factoring,

(2sinx-1)(sinx+1)=0

or sin(x)=1/2 or sin(x)=-1

Find all roots in the interval [0,2π].

- Trig. -
**mary**, Saturday, February 27, 2010 at 10:14pm
Thank you very much for the help

- Trig. :) -
**MathMate**, Sunday, February 28, 2010 at 8:16am
You are welcome!

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