Solving a 7th degree polynomial using De Moivre's theorem Use De Moivre Theorem to show that 
$$\cos 7θ=64\cos^7θ-112\cos^5θ+56\cos^3θ-7\cosθ$$ *Done
Hence obtain the roots of the equation 
$$128x^7-224x^5+112x^3-14x+1=0$$
in the form $\cos q\pi$
Attempt
$$\cos7θ=-1/2$$
$$θ=2π/21, 4π/21,8π/21,10π/21,14π/21,16π/21,20π/21$$
$$x=\cos θ$$
However the answer provided is
$\cos (\frac{2π}{21}+\frac{2kπ}{7})$ where $k=0,1,2,3,4,5,6$
Can somebody tell me what's wrong in my approach? 

You are correct.
When $k=0$, $\cos(\frac{2\pi}{21}+\frac{2k\pi}{7})=\cos\frac{2\pi}{21}$.
When $k=1$, $\cos(\frac{2\pi}{21}+\frac{2k\pi}{7})=\cos\frac{8\pi}{21}$.
When $k=2$, $\cos(\frac{2\pi}{21}+\frac{2k\pi}{7})=\cos\frac{14\pi}{21}$.
When $k=3$, $\cos(\frac{2\pi}{21}+\frac{2k\pi}{7})=\cos\frac{20\pi}{21}$.
When $k=4$, $\cos(\frac{2\pi}{21}+\frac{2k\pi}{7})=\cos\frac{26\pi}{21}=\cos\frac{16\pi}{21}$.
When $k=5$, $\cos(\frac{2\pi}{21}+\frac{2k\pi}{7})=\cos\frac{32\pi}{21}=\cos\frac{10\pi}{21}$.
When $k=6$, $\cos(\frac{2\pi}{21}+\frac{2k\pi}{7})=\cos\frac{38\pi}{21}=\cos\frac{4\pi}{21}$.
