August 6th, 2018, 09:11 AM  #1 
Newbie Joined: Aug 2018 From: Czech Republic Posts: 2 Thanks: 0  sqrt(2)*cos(2x)=sin(4x)
Hi, I've been struggling with this equation. I tried to expand cos(2x), but I just can't get to the result. Could someone please help me with this one? Thanks.

August 6th, 2018, 09:39 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,913 Thanks: 1113 Math Focus: Elementary mathematics and beyond 
it's not an identity. Did you type it correctly?

August 6th, 2018, 09:54 AM  #3  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,042 Thanks: 815 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Is the problem somehow 2 sin(2x) cos(2x) = sin(4x) ? This is an identity. Dan  
August 6th, 2018, 10:25 AM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,818 Thanks: 1463 
$\sqrt{2}\cos(2x) = \sin(4x)$ $\sqrt{2}\cos(2x) = 2\sin(2x)\cos(2x)$ $\sqrt{2}\cos(2x)  2\sin(2x)\cos(2x) = 0$ $\cos(2x){\large[}\sqrt{2}  2\sin(2x){\large]} = 0$ for $x \in [0, 2\pi )$ ... $\displaystyle\cos(2x) = 0 \implies x \in \left\{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right\}$ $\displaystyle\sin(2x) = \frac{\sqrt{2}}{2} \implies x \in \left\{ \frac{\pi}{8}, \frac{3\pi}{8}, \frac{9\pi}{8}, \frac{11\pi}{8} \right\}$ Last edited by skipjack; August 6th, 2018 at 02:40 PM. 

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