Solve cos^3x=cosxcos3x=cosx?

Cos^3x=cosxcos3x=cosx I get got the 0, pi/2, pi0,π2,π but I cant get (3pi)/23π2. If anyone can help out I would really appreciate it.

2 Answers
Feb 10, 2018

See below

Explanation:

It's a little tricky, but try setting the equation = 0

cos^3x=cosxcos3x=cosx

cos^3x-cosx=0cos3xcosx=0

Now factor out cosxcosx

cosx(cos^2x-1)=0cosx(cos2x1)=0

And now factor the difference of two squares

cosx(cosx+1)(cosx-1)=0cosx(cosx+1)(cosx1)=0

Set each factor equal to 0 and solve using your unit circle

1) cosx=0, x={pi/2, (3pi)/2}cosx=0,x={π2,3π2}

2) cosx+1=0cosx+1=0

cosx=-1, x=picosx=1,x=π

3) cosx-1=0cosx1=0
cosx=1, x=0cosx=1,x=0

Feb 11, 2018

See below.

Explanation:

cos^3(x)=cos(x)cos3(x)=cos(x)

cos^3(x)-cos(x)=0cos3(x)cos(x)=0

cos(x)(cos^2(x)-1)=0cos(x)(cos2(x)1)=0

cos(x)*(-1)(cos^2(x)-1)=0cos(x)(1)(cos2(x)1)=0

cos(x)(-cos^2(x)+1)=0cos(x)(cos2(x)+1)=0

Pythagorean identity:

cos(x)(sin^2(x))=0cos(x)(sin2(x))=0

:.

cos(x)=0=>pi/2,(3pi)/2

sin^2(x)=0=>0, pi,2pi