3cos2x=7cosx. Срочно пжжжжж

$3\cos 2x = 7\cos x \\ 3(2\cos ^{2}x - 1) - 7\cos x = 0 \\ 6\cos ^{2}x - 3 - 7\cos x = 0 \\ \cos x = t \\ 6t^{2}-7t-3=0 \\ D = 49 + 24*3 = 121 \\ \\ t_{1} = \dfrac{7 + 11}{12} = 1.5 \ ; \ \ \ t_{2} = \dfrac{7-11}{12} = -\dfrac{1}{3} \\ \\ $\left[ \begin{gathered} \cos x = 1.5 \\ \cos x = -\dfrac{1}{3} \\ \end{gathered} \right.$ \ \ \ ; \ $\left[ \begin{gathered} x \notin [-1;1] \\ x = \pm \arccos( -\dfrac{1}{3}) + 2\pi n, n \in Z \end{gathered} \right.$$

Ответ: x = arccos(-1/3) + πn, n∈Z