Find all solutions of the following equation in the interval [0,2pi). cos x +1= 2 sin^2 x Could anyone help me out?

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Answer 1 · 2017-11-03T12:00:16

Solution: In the interval #[0,2pi] x= pi/3 , x=pi , x=(5pi)/3#

#cosx +1 = 2sin^2x or 2sin^2x = cosx +1# or

#2(1-cos^2x)=cosx +1 or 2-2cos^2x=cosx +1# or

#2cos^2x+cosx +1-2=0 or 2cos^2x+cosx -1=0 # or

#2cos^2x +2cosx -cosx-1=0 # or

#2cosx(cosx+1)-1(cosx+1)=0# or

#(cosx+1)(2cosx-1)=0#.Either #cosx+1=0# or

#2cosx-1 =0#. Either #cosx=-1 or2cosx=1:. cosx =1/2#

When #cosx=-1 ; cos pi =-1 :. x = pi#

When #cosx=1/2 ; cos (pi/3) =1/2 and cos((5pi)/3) =1/2#

#:. x=pi/3 , x=(5pi)/3#.

Solution: In the interval #[0,2pi] x= pi/3 , x=pi , x=(5pi)/3# [Ans]