1. ## Pre Calc: Quad. Trig. Equations

So I went pretty smoothly through most of the homework questions untill I reached these ones. I'd love some quick help since I have a test on this tomorrow!

1) Find solutions on the interval [0, 2pi]:

4sinxtanx - 3tanx + 20sinx - 15 = 0

2) Use an identity to find solutions on the interval [0, 2pi]:

sec^2x - 2tan^2x = 0

Any hints/clues/help/ANSWERS would be greatly appreciated.

2. Originally Posted by mayflower29
2) Use an identity to find solutions on the interval [0, 2pi]:

sec^2x - 2tan^2x = 0

Any hints/clues/help/ANSWERS would be greatly appreciated.
remember $\sec^2 x = 1 + \tan^2 x$

3. Hello, mayflower29!

1) Find solutions on the interval $[0,\,2\pi]$
. . $4\sin x\tan x - 3\tan x + 20\sin x - 15 \:= \:0$

Factor: . $\tan x(4\sin x - 3) + 5(4\sin x - 3) \;=\;0$

Factor: . $(4\sin x-3)(\tan x + 5) \:=\:0$

Then: . $4\sin x-3 \:=\:0 \quad\Rightarrow\quad \sin x \:=\: \frac{3}{4} \quad\Rightarrow\quad x \:=\: \sin^{-1}\left(\frac{3}{4}\right) \;=\;\begin{Bmatrix}0.8481 \\ 2.2935\end{Bmatrix}$

. . . . . $\tan x+5 \:=\: 0 \quad\Rightarrow\quad \tan x \:=\:\text{-}5 \quad\Rightarrow\quad x \:=\:\tan^{-1}(\text{-}5)\;=\;\begin{Bmatrix}1.7682 \\ 4.9100\end{Bmatrix}$

4. Thanks a lot Soroban! That was perfect!

Originally Posted by Jhevon
remember $\sec^2 x = 1 + \tan^2 x$
Is it? I don't understand where that came from. We usually use our list of identities, and the only thing to do with secx is that its 1/cosx.

5. Originally Posted by mayflower29
Is it? I don't understand where that came from. We usually use our list of identities, and the only thing to do with secx is that its 1/cosx.
this is another identity, and it is the appropriate one to use here

6. Originally Posted by Jhevon
this is another identity, and it is the appropriate one to use here
Or you could convert the equation to sines and cosines and work from there.

$sec^2(x) - 2tan^2(x) = 0$

$\frac{1}{cos^2(x)} - \frac{2sin^2(x)}{cos^2(x)} = 0$

$\frac{1 - 2sin^2(x)}{cos^2(x)} = 0$

Now multiply both sides by $cos^2(x)$:
$1 - 2sin^2(x) = 0$

etc.

-Dan

7. Originally Posted by topsquark
Or you could convert the equation to sines and cosines and work from there.

$sec^2(x) - 2tan^2(x) = 0$

$\frac{1}{cos^2(x)} - \frac{2sin^2(x)}{cos^2(x)} = 0$

$\frac{1 - 2sin^2(x)}{cos^2(x)} = 0$

Now multiply both sides by $cos^2(x)$:
$1 - 2sin^2(x) = 0$

etc.

-Dan
Makes sense, thanks a lot!

8. Originally Posted by Soroban
Hello, mayflower29!

Factor: . $\tan x(4\sin x - 3) + 5(4\sin x - 3) \;=\;0$

Factor: . $(4\sin x-3)(\tan x + 5) \:=\:0$

Then: . $4\sin x-3 \:=\:0 \quad\Rightarrow\quad \sin x \:=\: \frac{3}{4} \quad\Rightarrow\quad x \:=\: \sin^{-1}\left(\frac{3}{4}\right) \;=\;\begin{Bmatrix}0.8481 \\ 2.2935\end{Bmatrix}$

. . . . . $\tan x+5 \:=\: 0 \quad\Rightarrow\quad \tan x \:=\:\text{-}5 \quad\Rightarrow\quad x \:=\:\tan^{-1}(\text{-}5)\;=\;\begin{Bmatrix}1.7682 \\ 4.9100\end{Bmatrix}$

Sorry, i was just wondering how did you get the second factor part you did.The (tanx+5 part) Can you please explain to me how you did that? Thankss

9. Originally Posted by aziz0917
Sorry, i was just wondering how did you get the second factor part you did.The (tanx+5 part) Can you please explain to me how you did that? Thankss
note that in the first line of what you quoted, the $4 \sin x - 3$ was a common term, so he factored it out. the result was that a factor of $\tan x + 5$ emerged