Question

So I have solve the following equations on the interval {0,2pi):

sin(x)cos(x)-sin(x)-cos(x)+1=0
sin(x-(pi/2))=cos(pi-x)

HELP I've spent three hours on countless youtube vids still cant find out how to find solutions :/

Answer 1

the first one:
I let (x) aside to shorten the typing
sincos -sin -cos +1=0
sin (cos-1) -(cos-1)=0 (factor sin out from the first 2 terms, let - sign out of the last 2 terms)
factor cos -1 out
(cos -1)(sin -1) 0
--> cos -1 =0 or sin -1 =0
you can handle from here, right?

Answer 2

the second one:
sin(x -pi/2) = sinx cos (pi/2) - sin(pi/2) cos x = -cos x
cos (pi-x) = cos pi cosx + sin pi sinx = cos x
so that, you have -cos x = cos x--> 2cos x =0 --> cos x =0 --> x = you can handle it, right?

Answer 3

Thanks a lot! I'm not sure what to do after I get cos-1=0 or sin-1=0 like what does evaluating it on the interval [0,2pi) mean exactly? How do I get solutions after getting that cos-1=0 and sin-1=0. Thanks for showing me how to factor btw!

Answer 4

\[\Large\rm \cos x-1=0\]Adding 1 to each side,\[\Large\rm \cos x=1\]You need to relate this back to your unit circle.Cosine is your horizontal component. (Movement along the x-axis).

So when cosx = 1, what angle does that put us at?