How do i solve secxcscx=2cscx? in terms of [0,2pi)

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Answer 1 · 2018-03-01T16:56:22

#x=pi/3, (5pi)/3#

Given:
#secxcscx=2cscx#

#secxcscx-2cscx=0#

#cscx(secx-2)=0#

cscx has its values more than 1 and less than -1 for x

Hence,

#secx-2=0#

#secx=2#

#secx=1/cosx#

#cosx=1/2#

#cos(pi/3)=1/2#

Between 0 and 2pi
cos is positive in first and fourth quadrants

#x=2npi+-pi/3#

#x=2(0)pi+pi/3, 2(1)pi-pi/3#
form the solution within the range

#x=pi/3, (5pi)/3#