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2tan x / 1-tan^2 x = 1/√3. What does x equal? a. 11pi/12 + n(pi) b. 5pi/12 + n(pi) c. 7pi/12 + n(pi) d. pi/12 + n(pi)
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\[\tan 2x=\frac{ 2 \tan x }{ 1-\tan ^2 x }\]
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No, I don't get any of this. I am so confused.
\[\tan 2x=\frac{ 1 }{ \sqrt{3} }=\tan \frac{ \pi }{ 6 }=\tan \left( n \pi+\frac{ \pi }{ 6 } \right)\],where n is an integer.
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