how do you solve -cosx=cos3x? for [0,2pi] thanx a lot:-)?

Question

i just dont remember how to do zis stuff

Answer 1

Ok, let's remember this identitie :

cos3x / cosx = 2cos2x - 1

cos3x = (2cos2x - 1)*cosx

-cosx = cos3x

-cosx = cosx*(2cos2x - 1)

cosx*(2cos2x - 1) + cosx = 0

cosx*(2cos2x) = 0

cos2x*cosx = 0

Now :

cos2x = 0 ; or cosx = 0

x E [ 0, 2pi]

2x = pi / 2, 3pi / 2 >>> x = pi / 4 and 3pi / 4

x = pi / 2, 3pi / 2

Those are the solutions

Hope that helps

Answer 2

cos 3x = cos(x+2x) = cos x cos 2x -sinx sin 2x=
cos x cos 2x -2sin^2 x cos x so
cosx(-1 -cos2x+2sin^2x)=0
so cos x = 0 and x= pi/2 and 2pi/2
(-1 -cos^2 x+3 -3cos ^2x)=0
-4 cos ^2x +2= 0
cos x= +-sqrt2 /2
x= pi/4,3pi/4,5pi/4 and 7pi/4