How do you solve #cos x cscx = 2cosx#?

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Answer 1 · 2016-04-01T12:25:01

For #ninZZ#

#x =2npi+pi/6 or x=2npi+(5pi)/6 or x = (2n+1)pi/2#

#cosx csc x = 2cosx#

#cosx 1/sinx=2cosx#

#cosx/sinx-2cosx=0#

#cosx(1/sinx-2)=0#

#cosx=0# or #1/sinx-2=0#

#cosx=0 => x = (2n+1)pi/2, ninZZ#

or

#1/sinx=2 => sinx = 1/2 #

#=> x =2npi+pi/6 or 2npi+(5pi)/6, ninZZ #