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

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

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

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.

To solve the given equation:

1. Start by factoring the equation if possible.

sin(theta)cos(theta) - sin(theta) - cos(theta) + 1 = 0

Rearranging terms:

sin(theta)(cos(theta) - 1) - (cos(theta) - 1) = 0

Now, notice that we have a common factor of (cos(theta) - 1):

(cos(theta) - 1)(sin(theta) - 1) = 0

2. To find the solution set, we need to consider the two cases where each factor equals zero.

Case 1: cos(theta) - 1 = 0

Solving this equation gives:

cos(theta) = 1

The solutions to this equation are when theta is 0 or 2π (since cosine is equal to 1 at these values).

Case 2: sin(theta) - 1 = 0

Solving this equation gives:

sin(theta) = 1

The solutions to this equation are when theta is π/2 or 3π/2 (since sine is equal to 1 at these values).

So, the solution set is:

θ = 0, π/2, 2π, 3π/2