How do you solve #(cosxcotx)/(1-sinx)=3# and find all solutions in the interval #[0,2pi)#?

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Answer 1 · 2016-09-06T14:05:30

#x = pi/6 and (5pi)/6#

#cosxcotx = 3(1 - sinx)#

#cosxcotx = 3 - 3sinx#

#cosx xx cosx/sinx = 3 - 3sinx#

#cos^2x/sinx = 3 - 3sinx#

#cos^2x = sinx(3 - 3sinx)#

#cos^2x = 3sinx - 3sin^2x#

#1 - sin^2x = 3sinx - 3sin^2x#

#3sin^2x - sin^2x - 3sinx + 1 = 0#

#2sin^2x - 3sinx + 1 = 0#

#2sin^2x - 2sinx - sinx + 1 = 0#

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

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

#sinx = 1/2 and sinx = 1#

#x = pi/6, (5pi)/6 and pi/2#

However, #pi/2# is extraneous, since it renders the denominator #0#.

Hopefully this helps!