Question

Find A when 0º≤A≤90º and 3tan2A = 2tanA + 1. (Enter only the number.)

Answer 1

\[\tan (2A) = \frac{2\tan A}{1-\tan^{2} A}\]
subbing that in
\[6 \tan A = (2\tan A+1)(1-\tan^{2}A)\]
\[2\tan^{3} A+\tan^{2}A+4\tan A-1 = 0\]
now you have to solve the cubic
\[2x^{3}+x^{2}+4x-1 = 0\]
http://www.wolframalpha.com/input/?i=2x%5E3%2Bx%5E2%2B4x-1+%3D+0

\[x = 0.23058\]
\[\tan A = 0.23058\]
\[A = \tan^{-1} (0.23058)\]

Answer 2

this seems more complicated than it should be...can you clarify i read your equation correctly?

is that tan^2 or tan(2A) ??