# Solve the following trig equation for all values on the interval of: 0 le x < 2pi sin4x=cos2x sin(2)(2x)=cos2x 2sinx2x=cos2x -cos2x -cos2x =2sin2x-cos2x=0 cos2x(2sin-1) cos2x=0 and 2sinx-1=0

Solve the following trig equation for all values on the interval of: $0\le x\le 2\pi$
$\mathrm{sin}4x=\mathrm{cos}2x$
$\mathrm{sin}\left(2\right)\left(2x\right)=\mathrm{cos}2x$
$2\mathrm{sin}x2x=\mathrm{cos}2x$
$-\mathrm{cos}2x-\mathrm{cos}2x$
$=2\mathrm{sin}2x-\mathrm{cos}2x=0$
$\mathrm{cos}2x\left(2\mathrm{sin}-1\right)$
$\mathrm{cos}2x=0$ and $2\mathrm{sin}x-1=0$
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yelashwag8
$\mathrm{sin}4x=\mathrm{cos}2x$
$\mathrm{sin}\left(2.2x\right)=\mathrm{cos}2x$
$2\mathrm{sin}\left(2x\right).\mathrm{cos}\left(2x\right)=\mathrm{cos}\left(2x\right)$
$2\mathrm{sin}\left(2x\right).\mathrm{cos}\left(2x\right)-\mathrm{cos}\left(2x\right)=0$
$\mathrm{cos}\left(2x\right).\left(2\mathrm{sin}\left(2x\right)-1\right)=0$

$\mathrm{cos}\left(2x\right)=\mathrm{cos}\left(3\pi /2\right)$
$2x=3\pi /2$
$x=3\pi /4$