How do you solve #sec^2(x) - sec(x) = 2#?

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Answer 1 · 2015-12-04T10:35:29

#x= \pi + 2k\pi#

#x = \pi/3 + 2k\pi#

# x= -pi\/3 + 2k\pi#

Since #sec(x)=1/cos(x)#, the expression becomes

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

Assuming #cos(x)\ne 0#, we can multiply everything by #cos^2(x)#:

#1-cos(x) = 2cos^2(x)#.

Rearrange:

#2cos^2(x)+cos(x)-1=0#.

Set #t=cos(x)#:

#2t^2+t-1=0#

Solve as usual with the discriminant formula:

#t=-1#, #t=1/2#

Convert the solutions:

#cos(x)=-1 \iff x=\pi+2k\pi#

#cos(x)=1/2 \iff x=\pm\pi/3 +2k\pi#