solve the equation


cos2x+2sinx=sin^2 x

I would appreciate a full explanation of how to get the correct answer

thanks

Write it all in terms of sine and then use a u sub to turn it into a quadratic.

Use the identities:

cos^{2}(x)=1-sin^{2}(x)

and

cos(2x)=cos^{2}(x)-sin^{2}(x)

Rewrite:

cos^{2}(x)-sin^{2}(x)+2sin(x)-sin^{2}(x)=0

1-sin^{2}(x)-sin^{2}(x)+2sin(x)-sin^{2}(x)=0

1-3sin^{2}(x)+2sin(x)=0

Now, let u=sin(x) and we get:

-3u^{2}+2u+1=0

3u^{2}-2u-1=0

Solve and resub.

First, break down your double angle into single angle:

cos2x + 2sinx = sin^2x

1-2sin^2x + 2sinx = sin^2x

Then, bring all the terms to a single hand side;

0=sin^2x-1+2sin^2x - 2sinx

0=3sin^2x- 2sinx -1

Factorise;

0=(3sin^2x- 1)(sinx +1)

Now, solve for;

0=3sin^2x- 1 and 0=sinx +1

Hope that helped! :)

LOL, I didn't know you were posting as well galactus :p

Hope that helped! :)[/QUOTE]

Yup , thanks:D

You're most welcomed millar2009S! :)