Asked by jh
Solve:
sin(theta)cos(theta)-sin(theta)−cos(theta)+1 = 0, 0 < or equal to theta < 2pi
Solution Set: ____________ ?
sin(theta)cos(theta)-sin(theta)−cos(theta)+1 = 0, 0 < or equal to theta < 2pi
Solution Set: ____________ ?
Answers
Answered by
Damon
sin(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
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
Answered by
Damon
sqrt (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
1-x^2 = 1
sin^2 T = 0
T = 0 or 180 degrees, 0 or pi
check 0 and pi in the original
Answered by
Damon
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.
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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