Official solution
\[f(x) = (4a - 3)(x + \ln 5) + 2(a - 7)\left( {{{\cos {x \over 2}} \over {\sin {x \over 2}}}.{{\sin }^2}{x \over 2}} \right)\]
\[f(x) = (4a - 3)(x + \ln 5) + (a - 7)\sin x\]
\[f'(x) = (4a - 3) + (a - 7)\cos x = 0\]
\[\cos x = {{ - (4a - 3)} \over {a - 7}}\]
\[ - 1 \le - {{4a - 3} \over {a - 7}} \le 1\]
\[ - 1 \le {{4a - 3} \over {a - 7}} \le 1\]
\[{{4a - 3} \over {a - 7}} - 1 \le 0\] and \[{{4a - 3} \over {a - 7}} + 1 \ge 0\]
\[ \Rightarrow {{ - 4} \over 3} \le a \le 2\]