How do you solve the equation #2cos^2theta-3costheta+1=0# for #0<=theta<2pi#?

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Answer 1 · 2016-11-21T11:29:07

The solutions are #S={0,pi/3,(5pi)/3} #

Let's factorise the LHS

#2cos^2theta-3costheta+1=(2costheta-1)(costheta-1)=0#

Therefore, #2costheta-1=0#

#costheta=1/2#

and #costheta-1=0#

#cos theta=1#

We are looking for #theta in [0, 2pi[#

When #costheta=1/2#

#theta=pi/3 , (5pi)/3#

When #costheta=1#

#theta=0#