1. ## Please send me in the right direction

Hey, I'm stuck on these problems if anyone could help me it would be amazing!

tan^2x=3/2secx

sqrt3 sinx - cosx=sqrt3

2. Originally Posted by progeek
Hey, I'm stuck on these problems if anyone could help me it would be amazing!

tan^2x=3/2secx

sqrt3 sinx - cosx=sqrt3
#1 ...

$\displaystyle \tan^2{x} = \sec^2{x} - 1$ ... solve the resulting quadratic for $\displaystyle \sec{x}$

#2 ...

$\displaystyle \sqrt{3} \cdot \sin{x} = \cos{x} + \sqrt{3}$

square both sides ...

$\displaystyle 3\sin^2{x} = \cos^2{x} + 2\sqrt{3} \cdot \cos{x} + 3$

$\displaystyle 3(1 - \cos^2{x}) = \cos^2{x} + 2\sqrt{3} \cdot \cos{x} + 3$

solve the quadratic for $\displaystyle \cos{x}$

3. Originally Posted by skeeter
#1 ...

$\displaystyle \tan^2{x} = \sec^2{x} - 1$ ... solve the resulting quadratic for $\displaystyle \sec{x}$

#2 ...

$\displaystyle \sqrt{3} \cdot \sin{x} = \cos{x} + \sqrt{3}$

square both sides ...

$\displaystyle 3\sin^2{x} = \cos^2{x} + 2\sqrt{3} \cdot \cos{x} + 3$

$\displaystyle 3(1 - \cos^2{x}) = \cos^2{x} + 2\sqrt{3} \cdot \cos{x} + 3$

solve the quadratic for $\displaystyle \cos{x}$

Thank you so much for the help. I got the second one but I'm still confused on the first problem. I got sec^2x-3/2secx-1. When I factor that out I get (sec-3/2)(sec+2/3) When I solve for both of those I don't get anything that I can redeem as an answer.

4. Originally Posted by progeek
Thank you so much for the help. I got the second one but I'm still confused on the first problem. I got sec^2x-3/2secx-1. When I factor that out I get (sec-3/2)(sec+2/3) When I solve for both of those I don't get anything that I can redeem as an answer.
$\displaystyle \sec^2{x} - \frac{3}{2}\sec{x} - 1 = 0$

$\displaystyle 2\sec^2{x} - 3\sec{x} - 2 = 0$

$\displaystyle (2\sec{x} + 1)(\sec{x} - 2) = 0$

5. Originally Posted by skeeter
$\displaystyle \sec^2{x} - \frac{3}{2}\sec{x} - 1 = 0$

$\displaystyle 2\sec^2{x} - 3\sec{x} - 2 = 0$

$\displaystyle (2\sec{x} + 1)(\sec{x} - 2) = 0$
ohhhhhhh, I feel so stupid. Yet at again thank you sooo much!