can someone solve the equation
cos4x=sin2x in the easiest way possible using trigometric formula. i keep going around in circles and getting no where.
$\displaystyle \cos^{2}2x-\sin^{2}2x-\sin2x=0$
$\displaystyle 1-\sin^{2}2x-\sin^{2}2x-\sin2x=0$
$\displaystyle 2\sin^{2}2x+\sin2x-1=0,\ \sin2x=t$
$\displaystyle 2t^2+t-1=0$
$\displaystyle \vdots$
$\displaystyle t_1=\frac12,\ t_2=-1$
$\displaystyle \ldots$
$1 - 2\sin^2(2x) = \sin(2x)$cos(4x)=sin(2x)
$0 = 2\sin^2(2x) + \sin(2x) - 1$
$0 = [2\sin(2x)-1][\sin(2x)+1]$
$\sin(2x) = \dfrac{1}{2}$
$2x = \dfrac{\pi}{6} \pm 2k\pi \implies x = \dfrac{\pi}{12} \pm k\pi \text{ where } k \in \mathbb{Z}$
$2x = \dfrac{5\pi}{6} \pm 2k\pi \implies x = \dfrac{5\pi}{12} \pm k\pi \text{ where } k \in \mathbb{Z}$
$\sin(2x) = -1$
$2x = \dfrac{3\pi}{2} \pm k \cdot 2\pi \implies x = \dfrac{3\pi}{4} \pm k \cdot \pi \text{ where } k \in \mathbb{Z}$