# How do you solve for x in 3sin2x=cos2x for the interval 0 ≤ x < 2π

Feb 13, 2015

You can use the fact that $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$
$3 \cdot \sin \left(2 x\right) = \cos \left(2 x\right)$
$3 \cdot \sin \frac{2 x}{\cos} \left(2 x\right) = 1$ and
$\tan \left(2 x\right) = \frac{1}{3}$
$\tan \left(2 x\right)$ is equal to $\frac{1}{3}$ for $2 x = 0.321 \mathmr{and} 3.463$ rad in your interval;
and so to have $x$ you divide by $2$ to get: $x = 0.160 \mathmr{and} 1.731$ rad