1. ## Algebraic Trig Solutions

"Find all solutions to $15sec(2x)sin(2x) + 9sec(2x) = 10sin(2x) + 6$ algebraically."

I am having trouble getting very far with this. Thanks.

Edit: I have tried messing around with the double argument formulas and turning them into different ratios.

2. Here's a start

$\sec(2x)\sin(2x) = \frac{\sin(2x)}{\cos(2x)} = \tan(2x)$

3. Hello, ATC!

Find all solutions to: . $15\sec2x\sin2x + 9\sec2x \:=\: 10\sin2x + 6$

Edit: I have tried messing around with the double argument formulas.
Since they are all double angles, why bother?

Multiply through by $\cos2x\!:$

. . $15\sin2x + 9 \;=\;10\sin2x\cos2x + 6\cos2x$

. . $10\sin2x\cos2x - 15\sin2x + 6\cos2x - 9 \;=\;0$

Factor: . $5\sin2x\bigg[2\cos2x - 3\bigg] + 3\bigg[\2\cos2x - 3 \bigg] \;=\;0$

Factor: . $(2\cos2x - 3)(5\sin2x + 3) \;=\;0$

And we have:

. . $2\cos2x - 3 \:=\:0 \quad\Rightarrow\quad \cos 2x \:=\:\frac{3}{2} \quad\hdots\;\text{ No real solutions.}$

. . $5\sin2x + 3 \:=\:0 \quad\Rightarrow\quad \sin2x \:=\:-\frac{3}{5}$

. . $2x \;=\;\arcsin(-0.6) \;=\; \begin{Bmatrix}\text{-}36.9^o + 360^on \\ 116.9^o + 360^on \end{Bmatrix}$

Therefore: . $x \;=\;\begin{Bmatrix}\text{-}18.45^o + 180^on \\ 58.45^o + 180^on \end{Bmatrix}\,\text{ for some integer }n.$

4. Thank you...multiplying by cos2x made it so much simpler.