The value of "x" that satisfies sec (x + 20°) = 2 ?

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Answer 1 · 2017-10-13T10:53:25

#x=n360^\circ+40^\circ or n360^\circ-80^\circ#

To make this easier to deal with, we will put each side under 1 because #sec(x+20^\circ)-=1/cos(x+20^\circ)#

#1/sec(x+20^\circ)=1/2#

#cos(x+20^\circ)=1/2#

#x+20^\circ=arccos(1/2)=60^\circ#

Actually though, #arccos(1/2)=n360^\circ+-60^\circ#

So, #x+20^\circ=n360^\circ+-60^\circ#

#x=n360^\circ+40^\circ or n360^\circ-80^\circ#

Answer 2 · 2017-10-13T10:55:05

Solution: # x=40^0 , x =-80^0#

#sec (x+20) =2 or 1/(cos (x+20))=2 # or

#cos (x+20) =1/2 #. We know # cos (+-60^0) =1/2#

#:. x+20 = +-60 :. x =60-20=40^0# or

#x = -60-20= -80^0#

Solution: # x=40^0 , x =-80^0# [Ans]