# Parallel tangent lines

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?

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

$$y(x)=\cos7x+7\cos x\to y'(x)=-7(\sin 7x+\sin x)$$
$$\sin 7x+\sin x=\sin \frac{7\pi}{6}+\sin \frac{\pi}{6}=0\\ 2\sin 4x \cos3x=0$$
• yup and I get the answer $x=k\frac{\pi}{4}$ and $x=\frac{\pi}{6}+k\frac{\pi}{3}$ – yolo expectz Mar 16 '17 at 21:32