1. ## Finding the value of x through factoring/quadratic?

School hasn't been going for a while and I really need a refresher. Our teacher gave us problems to tinker around abit and see how they are solved for. I've only listed half of what the teacher gave us since I've already got about to solving the other half.
I'm going to head out today and try to solve some of these later, since right now I still need to read up again.

Can you guys give me tips on how each one goes? Not too cryptic cause I'm really dull.
1. sqrt(3) * cot x * sin x + 2cos2 x = 0

2. 3tan2 x + 4sec x = -4

3. 10sin x2 - 7sin x = 6

Thanks in advance for those who'll give some tips. I'll check in again once I've went through all of these and answer (whether right or wrong) all or most of these.

2. ## Re: Finding the value of x through factoring/quadratic?

Originally Posted by RandomInquirer
School hasn't been going for a while and I really need a refresher. Our teacher gave us problems to tinker around abit and see how they are solved for. I've only listed half of what the teacher gave us since I've already got about to solving the other half.
I'm going to head out today and try to solve some of these later, since right now I still need to read up again.

Can you guys give me tips on how each one goes? Not too cryptic cause I'm really dull.
1. sqrt(3) * cot x * sin x + 2cos2 x = 0

2. 3tan2 x + 4sec x = -4

3. 10sin x2 - 7sin x = 6
For 1) you need $\displaystyle \cot(x)=\frac{\cos(x)}{\sin(x)}$ so $\displaystyle \cot(x)\sin(x)=\cos(x)$ gives an equation
$\displaystyle \sqrt3 y+2y^2=0$ to solve where $\displaystyle y=\cos(x)$.

For 2) $\displaystyle \tan^2(x)=\sec^2(x)-1$.

For 3) I suspect it should be $\displaystyle sin^2(x)$ and not $\displaystyle \sin(x^2)$

4. ## Re: Finding the value of x through factoring/quadratic?

Thanks for your help guys. I tried solving them but here's what I've come up with and for the most part if you guys don't mind I'll need your help again.

1. $\displaystyle (\sqrt3)\cot(x)\sin(x)+2cos^2(x)=0$
$\displaystyle (\sqrt3)\frac{\cos(x)}{\sin(x)}\sin(x)+2cos^2(x)=0$
I honestly don't know how to obtain the following: $\displaystyle \cot(x)\sin(x)=\cos(x)$
Won't it just be $\displaystyle (\sqrt3)\frac{\cos(x)}{\sin^2(x)}+2cos^2(x)=0$?
So I decided to try this out but then I ended up stuck in a rut.
$\displaystyle (\sqrt3)\frac{\cos(x)}{\1-cos^2(x)}+\frac{\2cos^2(x)}{1}=0$

$\displaystyle (\sqrt3)\frac{\cos(x)+[(2\cos^2(x))(\1-cos^2(x))]}{\1-cos^2(x)}=0$

$\displaystyle (\sqrt3)\frac{\cos(x)+2\cos^2(x)-2\cos^4(x)}{\1-cos^2(x)}=0$

I just froze here. I really don't know what to do now and I'm not sure if it's right.

2. I end up with $\displaystyle 3\sec^2(x) + 4\sec(x) + 3 = 0$
a = 3; b = 4; c = 3
Using the quadratic equation, I ended up with...
$\displaystyle \frac{(-4)±\sqrt(4^2-4(3)(3))}{2(3)} = \frac{(-4)±\sqrt(-20 )}{6}$
After doing the rest of the equation, then converting it to $\displaystyle cos^-$ in order for me to use the calculator then radicalizing the equation, I get the following:
x+ = $\displaystyle cos^-(\frac{\sqrt-20}{-4})$ and x- = $\displaystyle cos^-(\frac{3\sqrt-20}{-8})$
It doesn't seem right because of the negative square root and then I'm on the same boat I'm in at the first number.

3. I'm not sure about number three either, but here are my answers:
x+ = $\displaystyle \sin^-(1.2)$ = Math Error since 1.2 is beyond 1. >>>>>>>>>>(7 + 17)/20 = 1.2
x- = $\displaystyle \sin^-(-0.5)$ = -30 degrees >>>>>>>>>>>>>>>>>>>>>>>(7 - 17)/20 = -0.5

I have an additional question as well, since I ended up looking through other things.
Is is possible to solve $\displaystyle \sec(x) + \tan(x) = 1$ just by manipulating what's present?

5. ## Re: Finding the value of x through factoring/quadratic?

Originally Posted by RandomInquirer
Thanks for your help guys. I tried solving them but here's what I've come up with and for the most part if you guys don't mind I'll need your help again.

1. $\displaystyle (\sqrt3)\cot(x)\sin(x)+2cos^2(x)=0$
$\displaystyle (\sqrt3)\frac{\cos(x)}{\sin(x)}\sin(x)+2cos^2(x)=0$
I honestly don't know how to obtain the following: $\displaystyle \cot(x)\sin(x)=\cos(x)$
Won't it just be $\displaystyle (\sqrt3)\frac{\cos(x)}{\sin^2(x)}+2cos^2(x)=0$?
So I decided to try this out but then I ended up stuck in a rut.
$\displaystyle (\sqrt3)\frac{\cos(x)}{\1-cos^2(x)}+\frac{\2cos^2(x)}{1}=0$

$\displaystyle (\sqrt3)\frac{\cos(x)+[(2\cos^2(x))(\1-cos^2(x))]}{\1-cos^2(x)}=0$

$\displaystyle (\sqrt3)\frac{\cos(x)+2\cos^2(x)-2\cos^4(x)}{\1-cos^2(x)}=0$

I just froze here. I really don't know what to do now and I'm not sure if it's right.

2. I end up with $\displaystyle 3\sec^2(x) + 4\sec(x) + 3 = 0$
a = 3; b = 4; c = 3
Using the quadratic equation, I ended up with...
$\displaystyle \frac{(-4)±\sqrt(4^2-4(3)(3))}{2(3)} = \frac{(-4)±\sqrt(-20 )}{6}$
After doing the rest of the equation, then converting it to $\displaystyle cos^-$ in order for me to use the calculator then radicalizing the equation, I get the following:
x+ = $\displaystyle cos^-(\frac{\sqrt-20}{-4})$ and x- = $\displaystyle cos^-(\frac{3\sqrt-20}{-8})$
It doesn't seem right because of the negative square root and then I'm on the same boat I'm in at the first number.

3. I'm not sure about number three either, but here are my answers:
x+ = $\displaystyle \sin^-(1.2)$ = Math Error since 1.2 is beyond 1. >>>>>>>>>>(7 + 17)/20 = 1.2
x- = $\displaystyle \sin^-(-0.5)$ = -30 degrees >>>>>>>>>>>>>>>>>>>>>>>(7 - 17)/20 = -0.5

I have an additional question as well, since I ended up looking through other things.
Is is possible to solve $\displaystyle \sec(x) + \tan(x) = 1$ just by manipulating what's present?
You wondered "I honestly don't know how to obtain the following: $\displaystyle \cot(x)\sin(x)=\cos(x)$"
Youve already done it in the previous line:

$\displaystyle (\sqrt3)\frac{\cos(x)}{\sin(x)}\sin(x)+2cos^2(x)=0$

Simplify the sin(x) from the numerator with sin(x) in the denominator you will be left with cos(x). The sin(x) in front of the fraction is the same thing as sin(x) in the numerator. Also, you make the life easier for yourself if you make the change of variable cos(x) = y as suggested to you by Plato

Please put each question in a new post. Let's see how you will do.

Correction in previous post: cos(x) = 0 implies x = pi/2 or 3*pi/2, not 0 as I posted, you must have figured out.

6. ## Re: Finding the value of x through factoring/quadratic?

@votan: In your second to last post you mentioned that cos(x) = 0 implies x = 0. In fact cos(x) = 0 implies that x = pi/2, 3pi/2.

-Dan