# 6(tanx)^2-4(sinx)^2=1. Solve x?

Oct 11, 2017

$6 {\left(\tan x\right)}^{2} - 4 {\left(\sin x\right)}^{2} = 1$

$\implies 6 \left({\tan}^{2} x\right) \times {\cos}^{2} x - 4 {\sin}^{2} x \times {\cos}^{2} x = {\cos}^{2} x$

$\implies 6 {\sin}^{2} x - 4 {\sin}^{2} x \times {\cos}^{2} x = {\cos}^{2} x$

$\implies 6 \left(1 - {\cos}^{2} x\right) - 4 \left(1 - {\cos}^{2} x\right) {\cos}^{2} x = {\cos}^{2} x$

$\implies 6 - 6 {\cos}^{2} x - 4 {\cos}^{2} x + 4 {\cos}^{4} x = {\cos}^{2} x$

$\implies 4 {\cos}^{4} x - 11 {\cos}^{2} x + 6 = 0$

$\implies 4 {\cos}^{4} x - 8 {\cos}^{2} x - 3 {\cos}^{2} x + 6 = 0$

$\implies 4 {\cos}^{2} x \left({\cos}^{2} x - 2\right) - 3 \left({\cos}^{2} x - 2\right) = 0$

$= \left({\cos}^{2} x - 2\right) \left(4 {\cos}^{2} x - 3\right) = 0$

when ${\cos}^{2} x - 2 = 0 \implies \cos x \pm \sqrt{2}$

but $- 1 \le \cos x \le + 1$
so this solution is not acceptable.

when

$\left(4 {\cos}^{2} x - 3\right) = 0$

$\cos x = \pm \frac{\sqrt{3}}{2}$

For

$\cos x = + \frac{\sqrt{3}}{2} = \cos \left(\frac{\pi}{6}\right)$

$\implies x = 2 n \pi \pm \frac{\pi}{6} \text{ where } n \in \mathbb{Z}$

For

$\cos x = - \frac{\sqrt{3}}{2} = - \cos \left(\frac{\pi}{6}\right) = \cos \left(\frac{5 \pi}{6}\right)$

$\implies x = 2 n \pi \pm \frac{5 \pi}{6} \text{ where } n \in \mathbb{Z}$