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March 1st, 2016, 02:57 PM   #1
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Trig identities cos(x+3x)=sin4x/sin2x

Hey I found this problem, Can anyone solve this?

Cos(x+3x)=sin4x/sin2x
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March 1st, 2016, 04:24 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 $\cos(2x)$ ...

$2\cos^2(2x)-2\cos(2x)-1=0$

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

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 09:56 PM.
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