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).

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- Sep 10th 2008, 05:06 AMsanikuitrigo equation
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). - Sep 10th 2008, 05:19 AMmr fantastic
$\displaystyle \tan^3 A - \tan A = 0 \Rightarrow \tan A (\tan^2 A - 1) = 0$.

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. - Sep 10th 2008, 05:27 AMmr fantastic
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?) - Sep 10th 2008, 06:01 AMsanikui
question 1, i know do liaw....thx ~!

- Sep 10th 2008, 06:02 AMsanikui
O.o

For question 2, i know the answer A is 30*, 150*, 210*, 330*.

But the problem is didnt how to do.

Thx for helping~!V^^ - Sep 10th 2008, 01:28 PMmr fantastic
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?) - Sep 11th 2008, 04:46 AMsanikui
cos^2(A)=3/4 this undefined.

- Sep 11th 2008, 05:07 AMmr fantastic