# Thread: Another easy trig. problem...

1. ## Another easy trig. problem...

The equation is...

$\displaystyle 3sec^2\Theta-4tan\Theta=2$

I substituted $\displaystyle \dfrac{1}{cos^2\Theta}$ in place of $\displaystyle sec^2\Theta$ and $\displaystyle \dfrac{sin\Theta}{cos\Theta}$ in place of $\displaystyle tan\Theta$.

I now have...

$\displaystyle \dfrac{3}{cos^2\Theta}-\dfrac{4sin\Theta}{cos\Theta}=2$

I tried cross multiplying to get the left side as one fraction, but it doesn't seem to get me anywhere. It seems like there is some concept i am missing because i end up getting stuck in the same place with a lot of different problems like this.

note: i am 25 years old and not in high school or college. I am learning trig. because I want to.

Thanks!

2. Originally Posted by jonnygill
The equation is...

$\displaystyle 3sec^2\Theta-4tan\Theta=2$

I substituted $\displaystyle \dfrac{1}{cos^2\Theta}$ in place of $\displaystyle sec^2\Theta$ and $\displaystyle \dfrac{sin\Theta}{cos\Theta}$ in place of $\displaystyle tan\Theta$.

I now have...

$\displaystyle \dfrac{3}{cos^2\Theta}-\dfrac{4sin\Theta}{cos\Theta}=2$

I tried cross multiplying to get the left side as one fraction, but it doesn't seem to get me anywhere. It seems like there is some concept i am missing because i end up getting stuck in the same place with a lot of different problems like this.

note: i am 25 years old and not in high school or college. I am learning trig. because I want to.

Thanks!
$\displaystyle \sec^2{x} = 1 + \tan^2{x}$ ... sub for $\displaystyle \sec^2{x}$ and get a quadratic equation in terms of $\displaystyle \tan{x}$

3. thank you!

i used the quadratic equation and got 45 degrees and 18.435 degrees.