How do you find all values of x in the interval [0, 2pi] in the equation #cos(x+pi/3)+cosx=0#?

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Answer 1 · 2016-03-14T09:51:12

In domain #[0,2pi]#, solution is #{pi/3, (4pi)/3}#

We will use the identity #cos(x+y)=cosxcosy-sinxsiny#

Hence #cos(x+pi/3)+cosx=0#

or #cosxcos(pi/3)-sinxsin(pi/3)+cosx=0#

=or #cosx xx1/2-sinx xxsqrt3/2 +cosx=0#

or #sinx xxsqrt3/2=cosx xx3/2#

or #sinx/cosx=3/2xx2/sqrt3#

or #tanx=3/sqrt3=sqrt3#

or #x=pi/3#

But as #tan(pi+x)=tanx# another solution is #pi+pi/3=(4pi)/3#.

Hence, in domain #[0,2pi]#, solution is #{pi/3, (4pi)/3}#