Question #e9ec4

1 Answer
Fleur
Aug 3, 2017

#pi/3+(2pi)/3n, n in Z#

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 #cos x = y#.

#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 #0#:

#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 #pi/3#, #(5pi)/3#, and #pi# are all #(2pi)/3# apart. Thus, the solution can be written in simpler terms: #pi/3+(2pi)/3n, n in Z#.

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