Question #e31b3

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Answer 1 · 2016-09-23T13:47:26

First use the double angle formula you suggested and the identity $cottheta = 1/tantheta$

$3((2tanx)/(1 - tan^2x)) - 3/tanx = 0$

$(6(sinx/cosx))/(1 - sin^2x/cos^2x) - 3/(sinx/cosx) = 0$

$((6sinx)/cosx)/((cos^2x - sin^2x)/cos^2x) - (3cosx)/sinx = 0$

$(6sinxcosx)/(cos^2x - sin^2x) - (3cosx)/sinx = 0$

$(6sinxcosx)/(1 - sin^2x - sin^2x) - (3cosx)/sinx = 0$

$(6sinxcosx)/(1 - 2sin^2x) - (3cosx)/sinx = 0$

$6sin^2xcosx - 3cosx + 6cosxsin^2x = 0$

$cosx(6sin^2x - 3 + 6sin^2x) = 0$

$cosx(12sin^2x - 3) = 0$

$cosx(3(4sin^2x - 1)) = 0$

$cosx= 0" AND "sinx = +-1/2$

$x = pi/2, (3pi)/2, pi/6, (5pi)/6, (7pi)/6, (11pi)/6$

I have checked these solutions and they all seem to work.

Hopefully this helps!