# Thread: Factorable Trig equations (I'm totally lost with this one)

1. ## Factorable Trig equations (I'm totally lost with this one)

Okay.. I apologize for my inability to use the fancy math graphics. I'm going to finally put some time into learning that soon.

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Solve: 2 sec^2 (-)/2 + 3 sec (-)/2 - 2 = 0

provided 0 deg < (-) < 360 deg

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I've already tried a couple (most likely completely wrong) approaches... If someone could just walk me through this, I'd really appreciate it.

2. Originally Posted by Savior_Self
Okay.. I apologize for my inability to use the fancy math graphics. I'm going to finally put some time into learning that soon.

---

Solve: 2 sec^2 (-)/2 + 3 sec (-)/2 - 2 = 0

provided 0 deg < (-) < 360 deg

---

I've already tried a couple (most likely completely wrong) approaches... If someone could just walk me through this, I'd really appreciate it.
Do you mean $\displaystyle sec^2(\frac{\theta}{2})$?

This is a quadratic equation in $\displaystyle sec(\frac{\theta}{2})$. To make it easier substitute $\displaystyle u = sec(\frac{\theta}{2})$ to give

$\displaystyle 2u^2 + 3u - 2 = 0$

(2u-1)(u+2) = 0 so u = 0.5 or u = -2

ie: $\displaystyle sec(\frac{\theta}{2}) = 0.5$ in which case we can take the reciprocal of both sides to give us $\displaystyle cos(\frac{\theta}{2}) =2$. You'll get an error if you try solving so this is not a real solution.

$\displaystyle sec(\frac{\theta}{2}) = -2$

Again take the reciprocal to give $\displaystyle cos(\frac{\theta}{2}) = -0.5$
this gives $\displaystyle \frac{\theta}{2} = 120^o$ and $\displaystyle \theta = 240^o$

This is the only solution if I've plotted my graph right

3. Originally Posted by Savior_Self
Okay.. I apologize for my inability to use the fancy math graphics. I'm going to finally put some time into learning that soon.

---

Solve: 2 sec^2 (-)/2 + 3 sec (-)/2 - 2 = 0

provided 0 deg < (-) < 360 deg

---

I've already tried a couple (most likely completely wrong) approaches... If someone could just walk me through this, I'd really appreciate it.
$\displaystyle 2\sec^2{\frac{x}{2}} + 3\sec{\frac{x}{2}} - 2 = 0$

Let $\displaystyle X = \sec{\frac{x}{2}}$ so that the equation becomes

$\displaystyle 2X^2 + 3X - 2 = 0$

$\displaystyle 2X^2 + 4X - X - 2 = 0$

$\displaystyle 2X(X + 2) - (X + 2) = 0$

$\displaystyle (X + 2)(2X - 1) = 0$

$\displaystyle X + 2 = 0$ or $\displaystyle 2X - 1 = 0$

$\displaystyle X = -2$ or $\displaystyle X = \frac{1}{2}$

$\displaystyle \sec{\frac{x}{2}} = -2$ or $\displaystyle \sec{\frac{x}{2}} = \frac{1}{2}$.

Can you solve for $\displaystyle x$?

4. Thanks a ton to both of you. Turns out I was actually somewhat close, I just missed some critical steps that you've showed me. Thanks!