My Math Forum sqrt(2)*cos(2x)=sin(4x)

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 August 6th, 2018, 08:11 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, 08:39 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,945 Thanks: 1136 Math Focus: Elementary mathematics and beyond it's not an identity. Did you type it correctly?
August 6th, 2018, 08:54 AM   #3
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Quote:
 Originally Posted by ducanhle 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.
There are five solutions to the equation on $\displaystyle [ 0, ~2 \pi )$ which, I suspect, can only be found numerically.

Is the problem somehow 2 sin(2x) cos(2x) = sin(4x) ? This is an identity.

-Dan

 August 6th, 2018, 09:25 AM #4 Math Team     Joined: Jul 2011 From: Texas Posts: 2,947 Thanks: 1555 $\sqrt{2}\cos(2x) = \sin(4x)$ $\sqrt{2}\cos(2x) = 2\sin(2x)\cos(2x)$ $\sqrt{2}\cos(2x) - 2\sin(2x)\cos(2x) = 0$ $\cos(2x){\large[}\sqrt{2} - 2\sin(2x){\large]} = 0$ for $x \in [0, 2\pi )$ ... $\displaystyle\cos(2x) = 0 \implies x \in \left\{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right\}$ $\displaystyle\sin(2x) = \frac{\sqrt{2}}{2} \implies x \in \left\{ \frac{\pi}{8}, \frac{3\pi}{8}, \frac{9\pi}{8}, \frac{11\pi}{8} \right\}$ Thanks from greg1313, topsquark and ducanhle Last edited by skipjack; August 6th, 2018 at 01:40 PM.

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