Đáp án đúng: C
Giải chi tiết:\(\begin{array}{l}\,\,\,\,\,\cos 2x - {\tan ^2}x = \dfrac{{{{\cos }^2}x - {{\cos }^3}x - 1}}{{{{\cos }^2}x}}\\ \Leftrightarrow \dfrac{{\left( {2{{\cos }^2}x - 1} \right){{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x}} = \dfrac{{{{\cos }^2}x - {{\cos }^3}x - 1}}{{{{\cos }^2}x}}\\ \Leftrightarrow \left( {2{{\cos }^2}x - 1} \right){\cos ^2}x - {\sin ^2}x = {\cos ^2}x - {\cos ^3}x - 1\\ \Leftrightarrow {\cos ^2}x\left( {2{{\cos }^2}x - 1 + \cos x} \right) = 0\\ \Leftrightarrow 2{\cos ^2}x - 1 + \cos x = 0\,\,\left( {Do\,\,\cos x \ne 0} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = - 1\\\cos x = \dfrac{1}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \pi + k2\pi \\x = \dfrac{\pi }{3} + m2\pi \\x = - \dfrac{\pi }{3} + n2\pi \end{array} \right.\,\,\left( {k,m,n \in Z} \right)\end{array}\)
\(\begin{array}{l}\pi + k2\pi \in \left[ {1;70} \right] \Leftrightarrow 1 \le \pi + k2\pi \le 70\,\,\left( {k \in Z} \right) \Leftrightarrow \dfrac{{1 - \pi }}{{2\pi }} \le k \le \dfrac{{70 - \pi }}{{2\pi }}\,\,\left( {k \in Z} \right) \Leftrightarrow k \in \left\{ {0;1;2;...;10} \right\}\\\dfrac{\pi }{3} + m2\pi \in \left[ {1;70} \right] \Leftrightarrow 1 \le \dfrac{\pi }{3} + m2\pi \le 70\,\,\left( {m \in Z} \right) \Leftrightarrow \dfrac{{1 - \dfrac{\pi }{3}}}{{2\pi }} \le m \le \dfrac{{70 - \dfrac{\pi }{3}}}{{2\pi }}\,\,\left( {m \in Z} \right) \Leftrightarrow m \in \left\{ {0;1;2;...;10} \right\}\\ - \dfrac{\pi }{3} + n2\pi \in \left[ {1;70} \right] \Leftrightarrow 1 \le - \dfrac{\pi }{3} + n2\pi \le 70\,\,\left( {n \in Z} \right) \Leftrightarrow \dfrac{{1 + \dfrac{\pi }{3}}}{{2\pi }} \le n \le \dfrac{{70 + \dfrac{\pi }{3}}}{{2\pi }}\,\,\left( {n \in Z} \right) \Leftrightarrow n \in \left\{ {1;2;...;11} \right\}\end{array}\)
Vậy tổng các nghiệm thuộc \(\left[ {1;70} \right]\) của phương trình trên là :
\(\begin{array}{l}S = \left( {\pi + \pi + 2\pi + \pi + 4\pi + ... + \pi + 20\pi } \right) + \left( {\dfrac{\pi }{3} + \dfrac{\pi }{3} + 2\pi + \dfrac{\pi }{3} + 4\pi + ... + \dfrac{\pi }{3} + 20\pi } \right)\\\,\,\,\,\,\, + \left( { - \dfrac{\pi }{3} + 2\pi - \dfrac{\pi }{3} + 4\pi + .... - \dfrac{\pi }{3} + 22\pi } \right)\\S = 11\pi + \pi \left( {2 + 4 + ... + 20} \right) + \dfrac{{11\pi }}{3} + \left( {2 + 4 + ... + 20} \right)\pi - \dfrac{{11}}{3}\pi + \left( {2 + 4 + ... + 22} \right)\pi \\S = 11\pi + 110\pi + \dfrac{{11}}{3}\pi + 110\pi - \dfrac{{11}}{3}\pi + 132\pi \\S = 363\pi \end{array}\)
Chọn C.
