\(\begin{array}{l}
2)\,4{\cos ^2}\left( {x + 1} \right) - 1 = 0\\
\Leftrightarrow 4.\frac{{1 + \cos \left( {2x + 2} \right)}}{2} - 1 = 0\\
\Leftrightarrow 2\cos \left( {2x + 2} \right) + 1 = 0\\
\Leftrightarrow \cos \left( {2x + 2} \right) = - \frac{1}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x + 2 = \dfrac{{2\pi }}{3} + k2\pi \\
2x + 2 = - \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} - 1 + k\pi \\
x = - \dfrac{\pi }{3} - 1 + k\pi
\end{array} \right.\\
4)\,\,{\left( {\sin x - \cos x} \right)^2} - 1 = 0\\
\Leftrightarrow {\sin ^2}x - 2\sin x\cos x + {\cos ^2}x - 1 = 0\\
\Leftrightarrow 1 - \sin 2x - 1 = 0\\
\Leftrightarrow \sin 2x = 0\\
\Leftrightarrow 2x = k\pi \\
\Leftrightarrow x = \frac{{k\pi }}{2}
\end{array}\)
