find, for 0=<A=<360
(A:tan^(3A)=tanA)
i stuck at tan^3 and tanA.
find 6tan^2(A)-4sin^2(A)=1 , 0=<A=<360
i stuck at 6tan^2(A).
$\displaystyle \tan^3 A - \tan A = 0 \Rightarrow \tan A (\tan^2 A - 1) = 0$.
Case 1: $\displaystyle \tan A = 0$.
Case 2: $\displaystyle \tan^2 A - 1 = 0 \Rightarrow \tan^2 A = 1 \Rightarrow \tan A = \pm 1$.
Case 2 (a): $\displaystyle \tan A = 1$.
Case 2 (b): $\displaystyle \tan A = -1$.
Solve each case over the given domain.
One possible approach (there are many others):
Substitute $\displaystyle \tan^2 A = \frac{1}{\cos^2 A} - 1$ and $\displaystyle \sin^2 A = 1 - \cos^2 A$.
Re-arrange and simplify: $\displaystyle 2 \cos^4 A - 5 \cos^2 A + 3 = 0 \Rightarrow (\cos A - 1)(2 \cos A - 3) = 0$.
Therefore you need to solve $\displaystyle \cos A = 1$ over the given domain (the other case has no real solutions - why?)
I made several careless errors. Here is the corrected reply:
Re-arrange and simplify: $\displaystyle 4 \cos^4 A - 11 \cos^2 A + 6 = 0 \Rightarrow (\cos^2 A - 2)(4 \cos^2 A - 3) = 0$.
Therefore you need to solve $\displaystyle \cos^2 A = \frac{3}{4} \Rightarrow \cos A = \pm \frac{\sqrt{3}}{2}$ over the given domain (the other case has no real solutions - why?)