Trig problem?

Question

sin^2 x(x is not to the power as 2 is), anyways again "solve Sin^2 x.cos x=cos x for all values of x if x is measured in degrees"

Answer 1

Try to formulate the questio accurately.

If

Sin^2 x.cos x=cos x

means

((sin x)^2) * cos x = cos x

then the solution is x = 0°, x = 90°, x = 180°and x = 270°.

Answer 2

It's tempting to divide both sides by cos(x) right away, but we have to take into account the possibility that cos(x) = 0, which would give a solution. So one set of solutions is x = ±90, ±270, ±450, etc. (in the form of 90 ± 180n for integer n).

NOW that we have that case out of the way, divide both sides by cos(x). This gives sin^2 (x) = 1. So sin(x) = ±1. This gives c= 0, ±180, ±360, etc. as solutions.

Combining both, we get ±90n (where n is an integer) as a general solution.

Answer 3

((sin x)^2 )* (cos x) = cos x

If cos x is not equal to 0,

(sin x )^2 = 1
sin x = +1 and sin x = -1
x = (90 degrees +N*360) and x = (270 degrees + N*360), for N=0, 1, 2, ...