How to find the general equation of #2cos2theta=4costheta-3#?

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Answer 1 · 2018-06-05T11:23:13

#theta = pi/3+2npi# where n is an integer

#2cos(2theta)=4costheta-3#

#2(cos^2theta-sin^2theta)=4costheta-3#

#2(2cos^2theta-1)=4costheta-3#

#4cos^2theta-2=4costheta-3#

#4cos^2theta-4costheta+1=0#

#(2costheta-1)^2=0#

#2costheta-1=0#

#costheta=1/2#

#theta= cos^(-1)(1/2)#

#theta = pi/3+2npi# where n is an integer

Answer 2 · 2018-06-05T11:42:55

The general solution is #θ = 2 n pi ± 60#, where # n in Z#

Note: General solution instead of general equation is required

#2 cos 2 theta = 4 cos theta -3# or

#2 (2 cos^2 theta -1) = 4 cos theta -3# or

#4 cos^2 theta -2 = 4 cos theta -3# or

#4 cos^2 theta -2 -4 cos theta +3 =0# or

#4 cos^2 theta -4 cos theta +1 =0# or

#(2 cos theta -1)^2=0 :. 2 cos theta -1=0# or

#2 cos theta =1 or cos theta = 1/2 ; cos 60 =1/2#

and # cos (-60)=1/2 :. alpha= +-60^0#

The general solution is #θ =2 n pi ± 60#, where # n in Z# [Ans]