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March 31, 2015

March 31, 2015

Posted by **jh** on Thursday, March 25, 2010 at 7:33pm.

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 9:03pmsin(theta)cos(theta)-sin(theta)−cos(theta)+1 = 0

sinT(sqrt(1-sin^2T)) -sinT -sqrt(1-sin^2T)+1 = 0

let x = sinT

x sqrt(1-x^2) -x -sqrt(1-x^2) +1 = 0

(x-1) sqrt(1-x^2) = x-1

sqrt (1-x^2) = (x-1)/(x-1) =1

1-x^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 9:15pmsqrt (1-x^2) = (x-1)/(x-1) =1

1-x^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 9:18pmcheck 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.

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