solve cos^3x-3sin²xcosx?

Question

all i got was
cosx(cos²x-3sin²x)

Answer 1

continue from here.
cosx(1-sin²x-3sin²x) = 0
cosx(1-4sin²x) = 0
either cos x = 0 or sinx = 1/2 or sinx = -1/2
x = 0+2n*pi or x = -/+ pi/6 + n*pi

Answer 2

Do you mean :
cos³ x - 3sin² x cos x = 0 ?

cos x (cos² x - 3 sin² x) = 0
cos x (1 - sin² x - 3 sin² x) = 0 ( cos² = 1 - sin²)
cos x (1 - 4 sin² x) = 0
cos x (1 - 2 sin x) (1 + 2 sin x) = 0

This is true if :
cos x = 0 --> x = pi/2 + k.pi
sin x = 1/2 or x=-1/2 -> x = pi/6 + k.pi or -pi/6 + k.pi

with k = 0,1,2,3,....

Answer 3

You are right if you put cosx in factor than

cosx (cos^2x -3sin^2x) and sin^2 x= 1-cos^2x

cos x (cos^2x -3+ 3cos^2x)
cosx ( 4 cos^2x -3)

the roots of that equation are cos x =0 for x = (2n+1) pi/2

or cos x = (3/4 )^0.5 x = 30° +2nPi X= -30°+2nPI

Answer 4

cos^2 x- 3 sin ^2 x
= 1- sin^2 x - 3 sin^2 x
= 1 - 4 sin ^2 x

if you want to solve for = 0
then cos x = 0 or 1-4sin^2 x = 0

cos x = 0 => x = npi+pi/2

1- 4 sin ^2 x = 0 => sin x = 1/2 or - 1/2

sin x = 1/2 => x = pi/6 + 2npi or -pi/6+(2n+1)pi
sin x = -1/2 => x = 2npi - pi/6 or (2n+1) pi + pi/6

Answer 5

From here: cosx(cos^2x-3sin^2x) use the the rule cos^2x is related to sin^2x. I can't remember what it is, but it is in every trig book. What you want to do is change the sin^2x into terms of cosx. This probably doesn't make a whole bunch of sense, but hope it helps.