Question #fb8fe

1 Answer
Dec 2, 2017

#theta=pi/2,(3pi)/2#

Explanation:

#2cos^3theta-cos(3theta)=0#

We know:

#cos(3theta)=cos(2theta +theta)=cos(2theta)costheta-sin(2theta)sintheta#, and

#sin(2theta)=2sinthetacostheta#

#cos(3theta)=cos(2theta)costheta-2sinthetacosthetasintheta#

#cos(3theta)=cos(2theta)costheta-2costhetasin^2theta#

let's factor #costheta# out:

#cos(3theta)=costheta(cos(2theta)-2sin^2theta)#

Now let's plug this into the problem function:

#2cos^3theta-costheta(cos(2theta)-2sin^2theta)=0#

Let's factor out #costheta#

#costheta(2cos^2theta-cos(2theta)+2sin^2theta)=0#

#costheta(2(sin^2theta+cos^2theta)-cos(2theta))=0#

But we know:

#sin^2theta+cos^2theta=1#

#costheta(2-cos(2theta))=0#

Now, we set each term equal to zero and solve:

#costheta=0#, this gives us #theta=pi/2,(3pi)/2#

#2-cos(2theta)=0#, this gives us:

#cos(2theta)=2#

This is not acceptable because the value of a #cos# of any angle can not exceed #+-1#.