# trigo equation

• Sep 10th 2008, 05:06 AM
sanikui
trigo 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 AM
mr fantastic
Quote:

Originally Posted by sanikui
find, for 0=<A=<360
(A:tan^(3A)=tanA)

i stuck at tan^3 and tanA.

[snip]

$\displaystyle \tan^3 A - \tan A = 0 \Rightarrow \tan A (\tan^2 A - 1) = 0$.

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 AM
mr fantastic
Quote:

Originally Posted by sanikui
[snip]
find 6tan^2(A)-4sin^2(A)=1 , 0=<A=<360
i stuck at 6tan^2(A).

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 AM
sanikui
question 1, i know do liaw....thx ~!
• Sep 10th 2008, 06:02 AM
sanikui
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 PM
mr fantastic
Quote:

Originally Posted by mr 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?)

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 AM
sanikui
cos^2(A)=3/4 this undefined.
• Sep 11th 2008, 05:07 AM
mr fantastic
Quote:

Originally Posted by sanikui
cos^2(A)=3/4 this undefined.

No. It IS defined.