Another easy trig. problem...

**jonnygill** · February 17th 2011, 05:08 PM

The equation is...

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

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

I now have...

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

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!

**skeeter** · February 17th 2011, 05:40 PM

Originally Posted by jonnygill

The equation is...

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

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

I now have...

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

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!