# Thread: Finding solutions to this annoying equation ..

1. ## 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

2. $\sin(2x) = 2\sin(x)\cos(x)$

3. I did try that =\
This is all I got :

http://img8.imageshack.us/img8/9008/22102009064.jpg

4. 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.