How do you solve #cos(x-pi/6)=1+cos(x+pi/6)#?

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How do you solve #cos(x-pi/6)=1+cos(x+pi/6)#?

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Answer 1 · 2017-01-04T11:15:28

Solution : #x= ((3pi)/2) =270^0#

Explanation:

#cos(x-pi/6) = 1 +cos (x+pi/6) or cos(x-pi/6) - cos (x+pi/6) =1 or cos x cos(pi/6) - sin x sin (pi/6) - (cosxcos(pi/6) + sin x sin (pi/6)) =1 or cancel(cos x cos(pi/6)) - sin x sin (pi/6) - cancel(cosxcos(pi/6)) - sin x sin (pi/6) =1 or -2 sinx sin (pi/6) =1 or - cancel2*1/cancel2*sinx=1 # since #sin (pi/6)=1/2# or #sinx = -1 ; sin ((3pi)/2) = -1 :. sinx = sin ((3pi)/2) or x= ((3pi)/2) =270^0#[Ans]

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How do you solve #cos(x-pi/6)=1+cos(x+pi/6)#?

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