Решите уравнение: cos^2 (Пи/3 - 7х) = 1/2

$cos^2( \frac{\pi}{3} -7x)= \frac{1}{2} \\cos^2( \frac{\pi}{3} -7x)- \frac{1}{2} =0 \\(cos( \frac{\pi}{3} -7x)- \frac{\sqrt{2}}{2} )(cos( \frac{\pi}{3} -7x)+ \frac{\sqrt{2}}{2} )=0 \\cos(\frac{\pi}{3} -7x)- \frac{\sqrt{2}}{2}=0 \\cos(\frac{\pi}{3} -7x)=\frac{\sqrt{2}}{2} \\\frac{\pi}{3} -7x= \frac{\pi}{4} +2\pi n \\7x=\frac{\pi}{3}-\frac{\pi}{4}-2\pi n \\7x= \frac{\pi}{12} -2\pi n \\x_1= \frac{\pi}{84} - \frac{2\pi n}{7} ,\ n \in Z$
$\frac{\pi}{3} -7x=-\frac{\pi}{4} +2\pi n \\7x= \frac{\pi}{3} +\frac{\pi}{4} -2\pi n \\7x= \frac{7\pi}{12} -2\pi n \\x_2= \frac{\pi}{12} - \frac{2\pi n}{7},\ n \in Z \\cos( \frac{\pi}{3} -7x)+ \frac{\sqrt{2}}{2} =0 \\cos( \frac{\pi}{3} -7x)=-\frac{\sqrt{2}}{2} \\ \frac{\pi}{3} -7x= \frac{3\pi}{4} +2\pi n \\7x=\frac{\pi}{3}-\frac{3\pi}{4}-2\pi n \\7x=- \frac{5\pi}{12} -2\pi n \\x_3=- \frac{5\pi}{84} - \frac{2\pi n}{7} ,\ n \in Z$
$\frac{\pi}{3} -7x=-\frac{3\pi}{4} +2\pi n \\7x=\frac{\pi}{3}+ \frac{3\pi}{4} -2\pi n \\7x= \frac{13\pi}{12} -2\pi n \\x_4= \frac{13\pi}{84} - \frac{2\pi n}{7},\ n \in Z$
Ответ: $x_1= \frac{\pi}{84} - \frac{2\pi n}{7} ,\ n \in Z; \ x_2= \frac{\pi}{12} - \frac{2\pi n}{7},\ n \in Z; \ x_3=- \frac{5\pi}{84} - \frac{2\pi n}{7} ,\ n \in Z\\ x_4= \frac{13\pi}{84} - \frac{2\pi n}{7},\ n \in Z$