Given the following equations in the given interval 0 < x < 360

1 + tan^2x = 3tanx

i got x = -2.6 and x = -0.4
since the answers are negative, shudnt i use the 2nd and 4th quadrant to ge to get the answes? The answers at the back of the book use the 1st and 3rd quadrant. Have i gone wrong??????

2. Originally Posted by Sundae
Given the following equations in the given interval 0 < x < 360

1 + tan^2x = 3tanx

i got x = -2.6 and x = -0.4
since the answers are negative, shudnt i use the 2nd and 4th quadrant to ge to get the answes? The answers at the back of the book use the 1st and 3rd quadrant. Have i gone wrong??????
You can use symmetry to solve in any quadrant you wish.

Quad1: $\displaystyle \theta$

Quad2: $\displaystyle \pi-\theta$

Quad3: $\displaystyle \pi+\theta$

Quad4: $\displaystyle 2\pi-\theta$

3. Originally Posted by Sundae
Given the following equations in the given interval 0 < x < 360

1 + tan^2x = 3tanx

i got x = -2.6 and x = -0.4
since the answers are negative, shudnt i use the 2nd and 4th quadrant to ge to get the answes? The answers at the back of the book use the 1st and 3rd quadrant. Have i gone wrong??????
correction ...

$\displaystyle \tan^2{x} - 3\tan{x} + 1 = 0$

$\displaystyle \tan{x} = \frac{3 \pm \sqrt{5}}{2}$

$\displaystyle \tan{x} = 2.6180$

$\displaystyle \tan{x} = .382$

4. $\displaystyle x = .382$

5. Originally Posted by pickslides
$\displaystyle x = .382$
correction #2 ... $\displaystyle \tan{x} = .382$

6. thx alot, i know what i did wrong i missed off the - in the quadratic formula