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Find all solutions to the equation in radians: 2(sqrt3) - 4 = -4 - 2tan(theta/3)
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I can rewrite your equation as below: \[2\sqrt{3}=-2 \tan(\theta/3)\] so: \[\tan(\theta/3)=-\sqrt{3}\] whose solution are: |dw:1418835650359:dw| \[\frac{ \theta }{ 3 }=\frac{ 5 \pi }{ 3 }+k*\pi\] so: \[\theta=5 \pi+3*k*\pi\] where: \[k=0, \pm1, \pm2, \pm3....\]
I'm confused... where did the 5pi/3 come from? The unit circle?
yes! |dw:1418836685388:dw| you can choose: \[\frac{ \theta }{ 3 }=\frac{ 2 \pi }{ 3 }+k*\pi\] or \[\frac{ \theta }{ 3 }=\frac{ 5 \pi }{ 3 }+k*\pi\] as I did!
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