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March 1st, 2016, 03:57 PM   #1
Joined: Feb 2016
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Trig identities cos(x+3x)=sin4x/sin2x

Hey I found this problem, Can anyone solve this?

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March 1st, 2016, 05:24 PM   #2
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Your equation is not an identity ...

Left side is $\cos(4x)=2\cos^2(2x)-1$

Right side is $\dfrac{\sin(4x)}{\sin(2x)} =\dfrac{2\sin(2x) \cos(2x)}{\sin(2x)} = 2\cos(2x)$

results in a quadratic in $\cos(2x)$ ...


$\cos(2x)=\dfrac{1 - \sqrt{3}}{2}$

This last equation yields four solutions of $x$ for $0 < x < 2\pi$ ... I'll leave that part for you to calculate.

Last edited by skipjack; March 1st, 2016 at 10:56 PM.
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3xsin4x or sin2x, cosx, identities, trig

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