# Thread: Due tomorrow - solving trigonometric equations

1. ## 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!

2. Hello, littlelisa!

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

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

$\displaystyle 2\cos x\csc x - 4\cos x - \csc x + 2 \:=\:0$
Factor: .$\displaystyle 2\cos x(\csc x - 2) - (\csc x - 2) \;=\;0$

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

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

3. wow, i didn't even think of doing that, thanks so much! i really appreciate it!

4. Hello again, littlelisa!

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

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

And we have two equations to solve . . .

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

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

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

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

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