# tan²θ=-tanθ+3

• February 11th 2009, 01:19 AM
sarabara
tan²θ=-tanθ+3
thank you for any help!(Happy)
• February 11th 2009, 01:29 AM
geld
Do you know how to solve $y^2 = -y + 3$ ?
• February 11th 2009, 01:36 AM
$\tan^2\theta+\tan\theta-3=0$
Let $\tan\theta=x$ , thus $x^2+x-3=0$'
$x=\frac{-1+\sqrt{13}}{2}$
or $x=\frac{-1-\sqrt{13}}{2}$
Replace x with $\tan\theta$ , then solve for $\theta$