What are the All the Unit circle solutions to ---> (2 cosx)(sin^2) = (cosx)?

Answer 1

cosx(1-2sin^2x)=0
cosx(cos2x)=0
cosx=0
x=pi/2,3pi/2
cos2x=0
2x=pi/2 or 3pi/2
so x=pi/4 or 3pi/4

Answer 2

We have two cases:

Case 1, cosx not equal to zero - we can divide both sides by cosx:
2sin^2x = 1/2
sin x = 1/sqrt(2)
x=pi/4 or x=3pi/4.
EDIT: for got the negative root there, so you also have sinx=-1/sqrt(2) and two more solutions: x=5pi/4, x=7pi/4.

Case 2, cosx=0 - the equality holds. So x=pi/2 or x=3pi/2.

A total of 6 solutions.

Answer 3

Presuming you meant [sin(x)]^2 for the second term over there (since "sine squared" by itself doesn't make sense)

2cos(x) [sin(x)]^2 = cos(x)

Bring the cos(x) over to the left hand side

2cos(x) [sin(x)]^2 - cos(x) = 0

Factor

cos(x) [2 (sin(x))^2 - 1] = 0

cos(x) = 0
2 [sin(x)]^2 - 1 = 0

cos(x) = 0
[sin(x)]^2 = 1/2

cos(x) = 0
sin(x) = +/- 1/sqrt(2) (where sqrt means "square root")

Where is cos(x) equal to 0? It is equal to zero at two points: pi/2 and 3pi/2

Where is sin(x) equal to 1/sqrt(2)? At pi/4 and 3pi/4
Where is sin(x) equal to -1/sqrt(2)? At 5pi/4 and 7pi/4

So x = pi/2, 3pi/2, pi/4, 3pi/4, 5pi/4, 7pi/4