Due tomorrow - solving trigonometric equations

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• April 21st 2008, 08:50 PM
littlelisa
Due tomorrow - solving trigonometric equations
First time posting, hey guys!
I need urgent h/w help that's due tomorrow.

Here are the problems:

2cosxcscx - 4cosx - cscx + 2 (find all the answers in the intervals of 0 to 2pi)

5sin^23x + 8sin3x - 4
(i finally was able to factor this and then i don't know how to solve the equations from there)

sec^2xtan^2x + sec^2x = sec^4x
(verify identity i think)

Thanks!

LAST EDIT: i figured out the third one! :)
• April 21st 2008, 09:24 PM
Soroban
Hello, littlelisa!

Welcome aboard!
Here's the first one . . .

Quote:

Find all answer for: . $0 \,<\,x\,<\,2\pi$

$2\cos x\csc x - 4\cos x - \csc x + 2 \:=\:0$

Factor: . $2\cos x(\csc x - 2) - (\csc x - 2) \;=\;0$

Factor: . $(\csc x - 2)(2\cos x-1) \:=\:0$

And solve the two equations . . .
. . $\csc x - 2 \:=\:0\quad\Rightarrow\quad\csc x \:=\:2\quad\Rightarrow\quad x \:=\:\frac{\pi}{6},\;\frac{5\pi}{6}$
. . $2\cos x - 1 \:=\:0\quad\Rightarrow\quad\cos x \:=\:\frac{1}{2}\quad\Rightarrow\quad x \:=\:\frac{\pi}{6},\:\frac{5\pi}{3}$

• April 21st 2008, 09:26 PM
littlelisa
wow, i didn't even think of doing that, thanks so much! i really appreciate it!
• April 23rd 2008, 01:57 PM
Soroban
Hello again, littlelisa!

Quote:

$5\sin^2\!3x + 8\sin3x - 4 \:=\:0$ . . Solve for: $0 < x < 2\pi$

Factor: . $(\sin3x + 2)(5\sin3x - 2) \:=\:0$

And we have two equations to solve . . .

. . $\sin3x + 2 \:=\:0\quad\Rightarrow\quad \sin3x \:=\:-2\:{\color{red}?}$ . . . no real roots

. . $5\sin3x-2 \:=\:0\quad\Rightarrow\quad \sin3x \:=\:\frac{2}{5} \quad\Rightarrow\quad 3x \:=\:\sin^{-1}(0.4)$

. . An inverse trig function has many possible values . . .

. . . . . $3x \;=\;\begin{Bmatrix}0.42 \\ 2.73 \\ 6.69 \\ 9.01 \\ 12.98 \\ 15.30\end{Bmatrix}$

. . Therefore: . $x \;=\;\begin{Bmatrix}0.14 \\ 0.91 \\ 2.23 \\ 3.00 \\ 4.33 \\ 5.10 \end{Bmatrix}$