# Finding solutions to this annoying equation ..

• Oct 21st 2009, 03:52 PM
ZaZu
Finding solutions to this annoying equation ..
Find solutions for Sin(2x)-Sin(x)-Cos(x)+1=0

I have been trying to come up with my own ways but I keep failing =\

I tried squaring both sides after moving cos(x) to the other side ... I tried making 1 = Sin^2(x)+Cos^2(x) and solving it out .. But I just cant get it right !!!!

Any help please ?

Really appreciated :)
Thanks
• Oct 21st 2009, 04:03 PM
pickslides
$\sin(2x) = 2\sin(x)\cos(x)$
• Oct 21st 2009, 04:08 PM
ZaZu
I did try that =\
This is all I got :

http://img8.imageshack.us/img8/9008/22102009064.jpg
• Oct 21st 2009, 07:26 PM
mr fantastic
Quote:

Originally Posted by ZaZu
Find solutions for Sin(2x)-Sin(x)-Cos(x)+1=0

I have been trying to come up with my own ways but I keep failing =\

I tried squaring both sides after moving cos(x) to the other side ... I tried making 1 = Sin^2(x)+Cos^2(x) and solving it out .. But I just cant get it right !!!!

Any help please ?

Really appreciated :)
Thanks

$2 \sin x \cos x + 1 - (\sin x + \cos x) = 0$

Now substitute $1 = \cos^2 x + \sin^2 x$ and simplify:

$(\sin x + \cos x)^2 - (\sin x + \cos x) = 0$.

Factorise and then use the null factor law to get two simple trig equations. Solve them.