Solve (sinx-1)(cosx -1/2) = 0 where 0≤x<2pi

a) pi/3, pi/2, 5pi/3
b) pi/6, 5pi/6, pi
c) pi/3, pi, 5pi/3
d) pi/6, pi/2, 5pi/6

sin x = 1 when x = pi/2

cos x = 1/2 when x = pi/3 and -pi/3 which is 6 pi/3 - pi/3 = 5 pi/3

so
pi/3 , pi/2 , 5 pi/3 a)

and for heavens sake remember that
sin^2 x + cos^2 x = 1 !!!!

To solve the equation (sinx - 1)(cosx - 1/2) = 0, we need to find the values of x that make the equation true.

To do this, we set each factor equal to zero and solve for x.

1) sinx - 1 = 0

Adding 1 to both sides:
sinx = 1

Since the range of sin(x) is -1 to 1, the only solution to this equation is x = π/2.

2) cosx - 1/2 = 0

Adding 1/2 to both sides:
cosx = 1/2

To find the solutions, we can either use the unit circle or the inverse cosine function (cos^-1).

Using the unit circle, we can see that cos(x) = 1/2 when x = π/3 or x = 5π/3.

Using the inverse cosine function, we can find the solutions as follows:
x = cos^-1(1/2)
x = π/3 or x = 5π/3

Now, we have three possible values for x: π/2, π/3, and 5π/3.

Looking at the answer choices, we can see that the correct option is:

c) π/3, π, 5π/3