(2 điểm) Chứng minh dãy số \(\left( {{u_n}} \right)\) với \({u_n} = 1\,\,912 + n\,\,\,\left( {n \in {\mathbb{N}^*}} \right)\) là cấp số cộng. Xác định công sai, số hạng đầu của \(\left( {{u_n}} \right)\).

a) \[\sin 2x = \sin \frac{\pi }{5}\]

\( \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{5} + k2\pi \\2x = \pi - \frac{\pi }{5} + k2\pi \end{array} \right.\, \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{10}} + k\pi \\x = \frac{{2\pi }}{5} + k\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\).

b) \[\cot \left( {x + 50^\circ } \right) = \sqrt 3 \]

\( \Leftrightarrow \cot \left( {x + 50^\circ } \right) = \cot 30^\circ \)

\( \Leftrightarrow x + 50^\circ = 30^\circ + k180^\circ \)

\( \Leftrightarrow x = - 20^\circ + k180^\circ \,\,\left( {k \in \mathbb{Z}} \right)\).

c) \(\cos \left( {3x - \frac{\pi }{4}} \right) = \cos 2x\)

\( \Leftrightarrow \left[ \begin{array}{l}3x - \frac{\pi }{4} = 2x + k2\pi \\3x - \frac{\pi }{4} = - 2x + k2\pi \end{array} \right.\, \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{4} + k2\pi \\x = \frac{\pi }{{20}} + k\frac{{2\pi }}{5}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\).

d) \(\sin \left( {x + 6} \right) = 2{\cos ^2}5x - 1\)

\( \Leftrightarrow \sin \left( {x + 6} \right) = \cos 10x\)

\( \Leftrightarrow \sin \left( {x + 6} \right) = \sin \left( {\frac{\pi }{2} - 10x} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}x + 6 = \frac{\pi }{2} - 10x + k2\pi \\x + 6 = \pi - \frac{\pi }{2} + 10x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{22}} - \frac{6}{{11}} + k\frac{{2\pi }}{{11}}\\x = - \frac{\pi }{{18}} + \frac{2}{3} - k\frac{{2\pi }}{9}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\).

Theo bài ra ta có: \(\left\{ \begin{array}{l}{u_7} - {u_3} = - 8\,\,\,\left( 1 \right)\\{u_6} \cdot {u_9} = 72\,\,\,\,\,\left( 2 \right)\end{array} \right.\).

(1) \( \Leftrightarrow {u_1} + 6d - \left( {{u_1} + 2d} \right) = - 8 \Leftrightarrow d = - 2\).

(2) \( \Leftrightarrow \left( {{u_1} + 5d} \right)\left( {{u_1} + 8d} \right) = 72 \Leftrightarrow u_1^2 + 13d{u_1} + 40{d^2} = 72\)

\( \Leftrightarrow u_1^2 + 13 \cdot \left( { - 2} \right){u_1} + 40 \cdot {\left( { - 2} \right)^2} - 72 = 0 \Leftrightarrow u_1^2 - 26{u_1} + 88 = 0 \Leftrightarrow \left[ \begin{array}{l}{u_1} = 4\\{u_1} = 22\end{array} \right.\).