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**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?