Posted by jh on Thursday, March 25, 2010 at 7:33pm.
Solve:
sin(theta)cos(theta)sin(theta)−cos(theta)+1 = 0, 0 < or equal to theta < 2pi
Solution Set: ____________ ?

Precalculus(NEED HELP ASAP PLEASE!!)  Damon, Thursday, March 25, 2010 at 9:03pm
sin(theta)cos(theta)sin(theta)−cos(theta)+1 = 0
sinT(sqrt(1sin^2T)) sinT sqrt(1sin^2T)+1 = 0
let x = sinT
x sqrt(1x^2) x sqrt(1x^2) +1 = 0
(x1) sqrt(1x^2) = x1
sqrt (1x^2) = (x1)/(x1) =1
1x^2 = 1 or 1
sin^2 T = 0
or
sin^2 T = 2 which is impossible because sin T is between 1 and +1 
changed last lines  Damon, Thursday, March 25, 2010 at 9:15pm
sqrt (1x^2) = (x1)/(x1) =1
1x^2 = 1
sin^2 T = 0
T = 0 or 180 degrees, 0 or pi
check 0 and pi in the original 
check  Damon, Thursday, March 25, 2010 at 9:18pm
check T= 0 solution
sin(theta)cos(theta)sin(theta)−cos(theta)+1
0 (1)  0  1 + 1 = 0 check so zero works
check T = pi solution
0 0  (1) + 1 = 0
2 = 0 NO, so pi is not a solution.