Question #e9ec4
1 Answer
Aug 3, 2017
Explanation:
#sec x - 2 cos x = 1#
#sec x - 2 cos x -1 = 0#
#1/cos x - 2 cos x - 1 = 0#
#1/cos x - (2 cos^2 x)/cos x - cos x/cos x = 0#
#(1-2cos^2x - cos x) / cos x= 0#
#1-2cos^2x - cos x= 0#
#-2cos^2x - cos x + 1= 0#
#2cos^2x + cos x - 1= 0#
Let's say
#2y^2 + y - 1 = 0#
#2y^2 +2y - y - 1 = 0#
#2y(y+1) - 1(y+1) = 0#
#(2y-1)(y+1) = 0#
So, we can factor the equation above into
#(2 cos x - 1)(cos x +1)=0#
Set each of the factors equal to
#2 cos x - 1=0#
#2 cos x =1#
#cos x =1/2#
#color(green)(x=pi/3+2pin, x = (5pi)/3+2pin, n in Z)#
#cos x +1=0#
#cos x =-1#
#color(green)(x=pi + 2pin, n in Z)#
Notice that