Solve: tan²θ-(1+√3)tanθ+√3=0?

Jun 22, 2018

tan²θ-(1+√3)tanθ+√3=0

=>tan²θ-tantheta-√3tanθ+√3=0

=>tantheta(tanθ-1)-√3(tanθ-1)=0

=>(tanθ-1)(tantheta-√3)=0

So $\tan \theta = 1 = \tan \left(\frac{\pi}{4}\right)$

$\implies \theta = n \pi + \frac{\pi}{4} \text{ where } n \in \mathbb{Z}$

Again

$\tan \theta = \sqrt{3} = \tan \left(\frac{\pi}{3}\right)$

$\implies \theta = k \pi + \frac{\pi}{3} \text{ where } k \in \mathbb{Z}$

Jun 22, 2018

$t = \frac{\pi}{4} + k \pi$
$t = \frac{\pi}{3} + k \pi$

Explanation:

$f \left(t\right) = {\tan}^{2} t - \left(1 + \sqrt{3}\right) \tan t + \sqrt{3} = 0$.
Solve this quadratic equation for tan t.
Since (a + b + c = 0), use shortcut. The 2 real roots are:
tan t = 1 and $\tan t = \frac{c}{a} = \sqrt{3}$
a. $\tan t = 1$
Trig table and unit circle give --> $t = \frac{\pi}{4} + k \pi$
b. $\tan t = \sqrt{3}$
Trig table and unit circle give --> $t = \frac{\pi}{3} + k \pi$.
Check.
$t = \frac{\pi}{3}$ --> ${\tan}^{2} t = 3$ --> $- \left(1 + \sqrt{3}\right) \tan t = - \sqrt{3} - 3$.
$f \left(\frac{\pi}{3}\right) = 3 - \sqrt{3} - 3 + \sqrt{3} = 0$. Proved.
$t = \frac{\pi}{4}$ --> ${\tan}^{2} t = 1$ --> $- \left(1 + \sqrt{3}\right) \tan t = - 1 - \sqrt{3}$
$f \left(\frac{\pi}{4}\right) = 1 - 1 - \sqrt{3} + \sqrt{3} = 0$. Proved.