Question Part Points Submissions Used Use a Double or HalfAn

Question Part Points Submissions Used Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2). (Enter your answers as a comma-separated list.) cos(2) + cos() = 2 .Show all work done

Solution

cos(2) + cos() = 2

but   cos(2) =cos^2() -sin^2()

by identity sin^() +cos^2() =1

sin^2() = 1-cos^2()

so cos(2) = cos^2() -1 +cos^2() =2cos^2() -1

so now equation is

2cos^() - 1 +cos() =2

2cos^2() +cos() -1-2=0

2cos^() +cos() -3=0

2cos^2+3cos()-2cos() -3=0

cos() [2cos() +3] -1[2cos()+3]=0

(2cos()+3) (cos - 1)=0

so (2cos()+3) =0 or   (cos - 1)=0

  cos() =-3/2 or cos() =1

  cos() = -1.5 (not possoble since range of cosx is (-1,1) ),so ignore this

cos() =1

=0 is the only sloution

if interval is [0,2pi] then 2pi would have been a sloution also