1)
$\begin{array}{l}
\sin 5x - 2\sin x\left( {\cos 4x + \cos 2x} \right)\\
= \sin 5x - 2\sin x\cos 4x - 2\sin x\cos 2x\\
= \sin 5x - \left( {\sin 5x - \sin 3x} \right) - \left( {\sin 3x - \sin x} \right)\\
= \sin 5x - \sin 5x + \sin 3x - \sin 3x + \sin x\\
= \sin x
\end{array}$
2)
$\begin{array}{l} \cos x + \cos \left( {x + \dfrac{{2\pi }}{3}} \right) + \cos \left( {x - \dfrac{{2\pi }}{3}} \right)\\ = \cos x + \left[ {\cos \left( {x + \dfrac{{2\pi }}{3}} \right) + \cos \left( {x - \dfrac{{2\pi }}{3}} \right)} \right]\\ = \cos x + 2\cos \left( {\dfrac{{x + \dfrac{{2\pi }}{3} + x - \dfrac{{2\pi }}{3}}}{2}} \right)\cos \left( {\dfrac{{x + \dfrac{{2\pi }}{3} - x + \dfrac{{2\pi }}{3}}}{2}} \right)\\ = \cos x + 2\cos x\cos \dfrac{{2\pi }}{3}\\ = \cos x - 2\cos x.\cos \dfrac{\pi }{3}\\ = \cos x - 2.\cos x.\dfrac{1}{2} = \cos x - \cos x = 0 \end{array}$
