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August 6th, 2018, 08:11 AM   #1
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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.
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August 6th, 2018, 08:39 AM   #2
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it's not an identity. Did you type it correctly?
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August 6th, 2018, 08:54 AM   #3
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Quote:
Originally Posted by ducanhle View Post
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
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August 6th, 2018, 09:25 AM   #4
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$\sqrt{2}\cos(2x) = \sin(4x)$

$\sqrt{2}\cos(2x) = 2\sin(2x)\cos(2x)$

$\sqrt{2}\cos(2x) - 2\sin(2x)\cos(2x) = 0$

$\cos(2x){\large[}\sqrt{2} - 2\sin(2x){\large]} = 0$

for $x \in [0, 2\pi )$ ...

$\displaystyle\cos(2x) = 0 \implies x \in \left\{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right\}$

$\displaystyle\sin(2x) = \frac{\sqrt{2}}{2} \implies x \in \left\{ \frac{\pi}{8}, \frac{3\pi}{8}, \frac{9\pi}{8}, \frac{11\pi}{8} \right\}$
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Last edited by skipjack; August 6th, 2018 at 01:40 PM.
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