sin(x+pi)+cos(x-(pi/2))=1 HELP! Find x?

Question

find values of X in domain [0,2pi]

sin(x+pi)+cos(x-(pi/2))=1
this is what I did. is the 1st circle ze right thing?
http://i3.tinypic.com/53p67vo.jpg

the right answer is:
x=pi/6 ; 2pi/3 ; 7pi/6 ; 5pi/3
but how do I get to it???

Answer 1

sin(x + pi) = sinx cospi + sinpi cosx = -sinx

cos(x -pi/2) = cosx cos(pi/2) + sinx sin(pi/2) = sinx

-sinx + sinx = 0 = 1 there is a problem !!

Answer 2

as somebody else pointed out

sin(x+pi)+cos(x-(pi/2)) reduces to -sinx + sinx =0

there might be some error in the way you wrote the question

as for your method
your formula for cos(A-B) is wrong
since
cos(A-B) = cosA cosB + sinA sinB

also you cannot separate sin or cos from sin(x) or cos(x)
it is a function like f(x)
therefore your method is wrong

you should be using the values of sin(pi) , sin(pi/2)
cos(pi) ,cos(pi/2) etc after expanding, if you start off the way you did.