# Another easy trig. problem...

• Feb 17th 2011, 05:08 PM
jonnygill
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!
• Feb 17th 2011, 05:40 PM
skeeter
Quote:

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}\$
• Feb 17th 2011, 07:29 PM
jonnygill
thank you!

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