The question: For which values of $x$ will the tangent line of $y(x)=\cos7x+7\cos x$ at a point with $x=x$ be parallel to the tangent line at a point with $x=\frac{\pi}{6}$

So the 2 tangent lines will be parallel they are equal so $y'(x)=-7\sin7x-7\sin x$ $-7\sin7x-7\sin x=-7\sin7\frac{\pi}{6}-7\sin\frac{\pi}{6}$ by my calculations $-7\sin7x+7\sin x=0$ which I can't simplify further.

My question: Can you find the value of $x$ and is this the right idea?

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    $\begingroup$ I have missed the application of the chain rule $\frac{d}{dx}( \cos 7x) = -7\sin 7x$ $\endgroup$ – Doug M Mar 16 '17 at 21:11

$$y(x)=\cos7x+7\cos x\to y'(x)=-7(\sin 7x+\sin x)$$


$$\sin 7x+\sin x=\sin \frac{7\pi}{6}+\sin \frac{\pi}{6}=0\\ 2\sin 4x \cos3x=0$$

Can you finish?

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    $\begingroup$ Oh, forgot about the sum-to-product identity. Good catch. $\endgroup$ – Matthew Leingang Mar 16 '17 at 21:22
  • $\begingroup$ yup and I get the answer $x=k\frac{\pi}{4}$ and $x=\frac{\pi}{6}+k\frac{\pi}{3}$ $\endgroup$ – yolo expectz Mar 16 '17 at 21:32

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