How to solve the equation? 2cosθ/2+1=0 0<θ<2π

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Answer 1 · 2018-03-28T06:55:54

#theta=(4pi)/3#

We are given

#2cos(theta/2)+1=0#

#rArr2cos(theta/2)=-1#

#rArrcos(theta/2)=-1/2#

#rArrtheta/2=arccos(-1/2)#

This will return 2 possible values for #theta/2#. They are:

#rArrtheta/2=(2pi)/3, (4pi)/3#

#rArrtheta=(4pi)/3, (8pi)/3#

However, the question states that #0##<##theta<2pi#

So our only valid solution is #theta=(4pi)/3#