How do you find the exact value of #2cosx-secx=0# in the interval #0<=x<360#?
1 Answer
Nov 23, 2016
#2cosx - 1/cosx = 0#
#2cos^2x - 1 = 0#
#cos^2x= 1/2#
#cosx= +-1/sqrt(2)#
#x = 45˚, 135˚, 225˚, 315˚#
Hopefully this helps!
