# Thread: Solve each equation for 0 < theta < 2(pie)

1. ## Solve each equation for 0 < theta < 2(pie)

Solve each equation for 0 < theta < 2(pie)

tan^2(theta) + 2tan(theta) - 5 = 0

I can't even remember how to do this...a little help? Please and thanks!

2. Originally Posted by Jeavus
Solve each equation for 0 < theta < 2(pie)

tan^2(theta) + 2tan(theta) - 5 = 0

I can't even remember how to do this...a little help? Please and thanks!
it is quadratic in $\tan \theta$ so treat it as you would a quadratic equation

3. I'm sorry Jhevon, but I just don't know how to approach it.

I've tried using the identity tan(theta) = sin(theta)/cos(theta), but I can't set it up.

4. I could substitue tan(theta) for x to make it:

x^2 + 2x - 5 = 0

Would I have to use to solve for the roots?

5. Originally Posted by Jeavus
I could substitue tan(theta) for x to make it:

x^2 + 2x - 5 = 0

Would I have to use to solve for the roots?
yes, that's what i meant it was quadratic in tan. you have to remember to replace x with tan(theta) when done though to find theta

6. When subbing in for 4ac though...you get a negative number in the square root.

What then?

7. Originally Posted by Jeavus
When subbing in for 4ac though...you get a negative number in the square root.

What then?
no, the number is positive. you have a negative times a negative times a positive, the result is positive

8. Hmm...I'm still not getting the answers it is giving:

0.97, 1.85, 4.11, 5.00 (in the textbook)

After doing calculations, the 2 roots are:

x = 0.449
x2 = 4.449

I convert both xs back into tan(theta).

tan(theta1) = 0.449
tan(theta2) = 4.449

Solve for both degree.

tan(theta1) = 24.2 degrees
tan(theta2) = 77.3 degrees

theta1 = 0.422
theta 2 = 1.34

Which do not resemble the answers in the textbook. I'm lost.

9. Originally Posted by Jeavus
Hmm...I'm still not getting the answers it is giving:

0.97, 1.85, 4.11, 5.00 (in the textbook)

After doing calculations, the 2 roots are:

x = 0.449
x2 = 4.449

I convert both xs back into tan(theta).

tan(theta1) = 0.449
tan(theta2) = 4.449

Solve for both degree.

tan(theta1) = 24.2 degrees
tan(theta2) = 77.3 degrees

theta1 = 0.422
theta 2 = 1.34

Which do not resemble the answers in the textbook. I'm lost.
you solved for the x's wrong somehow

10. x = (-b)+/- ((sqroot)(2^2 - 4(1)(-5)))/2)1)

Is that not correct?

11. Originally Posted by Jeavus
x = (-b)+/- ((sqroot)(2^2 - 4(1)(-5)))/2)1)

Is that not correct?
we have $x = \tan \theta = \frac {-2 \pm \sqrt{2^2 - 4(1)(-5)}}2$

now simplify and solve for $\theta$ (remember to use referefernce angles to get all the angles in $0 \le \theta \le 2 \pi$. as the answer tells you, you should have 4 angles not just two