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\frac{\left(\sin 2x+sin 4x\right)}{\left(sin 2x - sin 4x\right)}+\frac{tan 3x}{tan x}=0

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Final answer to the problem

$\frac{\left(2\left(x\cos\left(x\right)+\sin\left(x\right)\right)+4\cos\left(4x\right)\right)\left(\sin\left(2x\right)-\sin\left(4x\right)\right)+\left(-2x\sin\left(x\right)-\sin\left(4x\right)\right)\left(2\cos\left(2x\right)-4\cos\left(4x\right)\right)}{\left(\sin\left(2x\right)-\sin\left(4x\right)\right)^2}+3\sec\left(3x\right)^2=0$

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Math interpretation of the question

$\frac{d}{dx}\left(\frac{2x\sin\left(x\right)+\sin\left(4x\right)}{\sin\left(2x\right)-\sin\left(4x\right)}+\tan\left(3x\right)=0\right)$

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$\frac{d}{dx}\left(\frac{2x\sin\left(x\right)+\sin\left(4x\right)}{\sin\left(2x\right)-\sin\left(4x\right)}+\tan\left(3x\right)=0\right)$

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Learn how to solve problems step by step online. \frac{\left(\sin 2x+sin 4x\right)}{\left(sin 2x - sin 4x\right)}+\frac{tan 3x}{tan x}=0. Math interpretation of the question. Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable. The derivative of the constant function (0) is equal to zero. The derivative of a sum of two or more functions is the sum of the derivatives of each function.

Final answer to the problem

$\frac{\left(2\left(x\cos\left(x\right)+\sin\left(x\right)\right)+4\cos\left(4x\right)\right)\left(\sin\left(2x\right)-\sin\left(4x\right)\right)+\left(-2x\sin\left(x\right)-\sin\left(4x\right)\right)\left(2\cos\left(2x\right)-4\cos\left(4x\right)\right)}{\left(\sin\left(2x\right)-\sin\left(4x\right)\right)^2}+3\sec\left(3x\right)^2=0$

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