De Moivre's formula to solve $\cos$ equation Use De Moivre’s formula to show that
$$
\cos\left(x\right) + \cos\left(3x\right) = 2\cos\left(x\right)\cos\left(2x\right)
$$
$$
\mbox{Show also that}\quad
\cos^{5}\left(x\right) = \frac{1}{16}\,\cos\left(5x\right) + \frac{5}{16}\,\cos\left(3x\right) + \frac{5}{8}\,\cos\left(x\right)
$$
Hence solve the equation $\cos\left(5x\right) = 16\cos^{5}\left(x\right)$ completely. 
Express your answers as rational multiples of $\pi$.

@Lauren, the approach comes down to the fact that $Re[(\text{cos}(x)+i\cdot \text{sin}(x))^3] = \text{cos}(3x)$.
So, $$\text{cos}(3x) \\ = \text{cos}^3(x) + 3\text{cos}(x)\cdot (i^2\cdot \text{sin}^2(x)) \\ = \text{cos}(x)\cdot (\text{cos}^2(x) - 3\text{sin}^2(x)) \\ = \text{cos}(x)\cdot (1-2\text{sin}^2(x)+ \text{cos}(2x)-1) \\ = \text{cos}(x)\cdot (2\text{cos}(2x) - 1) = \text{cos}(3x).$$
This proves part 1. Then the next part follows from the fact that $\text{cos}(3x) = -2\text{cos}(x)$. Why? I'll leave this as an exercise.
