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millar2009S · May 20, 2009, 11:18 AM

solve the equation

cos2x+2sinx=sin^2 x

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

thanks

galactus · May 20, 2009, 11:55 AM

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.

Unknown008 · May 20, 2009, 11:59 AM

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! :)

Unknown008 · May 20, 2009, 12:00 PM

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

millar2009S · May 20, 2009, 12:12 PM

Hope that helped! :)[/QUOTE]

Yup , thanks:D

Unknown008 · May 20, 2009, 12:14 PM

You're most welcomed millar2009S! :)