Question #9c5a0

May 5, 2016

$x = \frac{\pi}{4} + \pi k$, $k \in \mathbb{Z}$

Explanation:

$1 + {\sin}^{2} \left(x\right) = 3 \sin \left(x\right) \cos \left(x\right)$, $\tan \left(x\right) \ne \frac{1}{2}$

$\implies \left({\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right)\right) + {\sin}^{2} \left(x\right) = 3 \sin \left(x\right) \cos \left(x\right)$

$\implies {\cos}^{2} \left(x\right) + 2 {\sin}^{2} \left(x\right) = 3 \sin \left(x\right) \cos \left(x\right)$

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

$\implies \left(\cos \left(x\right) - \sin \left(x\right)\right) \left(\cos \left(x\right) - 2 \sin \left(x\right)\right) = 0$

$\implies \cos \left(x\right) - \sin \left(x\right) = 0$ or $\cos \left(x\right) - 2 \sin \left(x\right) = 0$

As the second equation may be expressed as $\sin \frac{x}{\cos} \left(x\right) = \frac{1}{2}$, violating our initial condition of $\tan \left(x\right) \ne \frac{1}{2}$, we will only focus on the first.

$\cos \left(x\right) - \sin \left(x\right) = 0$

$\implies \cos \left(x\right) = \sin \left(x\right)$

$\therefore x = \frac{\pi}{4} + \pi k$, $k \in \mathbb{Z}$