how do you solve sin^2(x)cos(x)=cos(x)?

Answer 1

sin^2(x) = 1

sin(x) = ±1

x = π/2, 3π/2, ...

Answer 2

sin^2(x) * cos(x) = cos(x)
sin^2(x) = 1 [divide both sides by cos(x) ]
sin (x ) = 1 [take the square root of both sides ]
when does the sine function equal 1? Well, it is at 0 initially, and at 0 again when x = pi. The sine function is bounded at y = 1, therefore, it is at its max of 1 halfway between 0 and pi (it is best to look at the graph) and that is at pi/2

Answer 3

Here you will have to make an assumption. The assumption is that cos(x) <> 0
then
sin^2(x).cos(x)=cos(x)
is equivalent to
sin^2(x)=1
or,
sin(x) =+1
or sin(x)=-1
therefore,
x=(2i+1)*90 degrees for all i in integer set including zero
The other solution is
cos(x)=0
i.e.,
x=(2i+1)*90 degrees for all i in integer set including zero. Both the results are identical

Answer 4

sin^2(x)cos(x)=cos(x)
sin^2(x) = 1
sin (x) = +/- 1
x = arcsin 1 = pi/2 radians = 90 degrees
x = arcsin (-1) = -pi/2 radians = -90 degrees, and
x = 3pi/2 = 270 degrees.

These values repeat every 2pi.